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Q1.

In the above example the angular velocity of $\mathrm { S } _ { 2 }$ as actually observed by an astronaut in $\mathrm { S } _ { 1 }$ is -

$\pi / 3$ rad/hr

B

$\pi / 3$ rad/sec

C

$\pi / 6 \mathrm { rad } / \mathrm { hr }$

D

2$\pi / 7$ rad/hr

##### Explanation

When $\mathrm { S } _ { 2 }$ is closest to $\mathrm { S } _ { 1 } ,$ the speed of $\mathrm { S } _ { 2 }$ relative to $\mathrm { S } _ { 1 }$is $\mathrm { v } _ { 2 } - \mathrm { v } _ { 1 } = \pi \times 10 ^ { 4 } \mathrm { km } / \mathrm { hr }$ . The angular speed of $\mathrm { S } _ { 2 }$ asobserved from $\mathrm { S } _ { 1 }$ (when closest distance between themis $\mathrm { r } _ { 2 } - \mathrm { r } _ { 1 } = 3 \times 10 ^ { 4 } \mathrm { km }$ )
$\omega = \frac { \mathrm { v } _ { 2 } - \mathrm { v } _ { 1 } } { \mathrm { r } _ { 2 } - \mathrm { r } _ { 1 } } = - \frac { \pi \times 10 ^ { 4 } } { 3 \times 10 ^ { 4 } } = - \frac { \pi } { 3 } \mathrm { rad } / \mathrm { hr }$
$| \omega | = \frac { \pi } { 3 } \mathrm { rad } / \mathrm { hr }$

Q2.

The moon revolves round the earth 13 times in one year. Ifthe ratio of sun-earth distance to earth-moon distance is $392 ,$ then the ratio of masses of sun and earth will be -

A

365

B

356

$3.56 \times 10 ^ { 5 }$

D

1

##### Explanation

Period of revolution of earth around sun
$\mathrm{T}_{\mathrm{e}}^{2}=\frac{4 \pi^{2} \mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{GM}_{\mathrm{s}}}$
Period of revolutions of moon around earth
$\mathrm{T}_{\mathrm{n}}^{2}=\frac{4 \pi^{2} \mathrm{R}_{\mathrm{m}}^{2}}{\mathrm{GM}_{\mathrm{e}}}$
$\therefore \left( \frac { \mathrm { T } _ { \mathrm { e } } } { \mathrm { T } _ { \mathrm { m } } } \right) ^ { 2 } = \left( \frac { \mathrm { M } _ { \mathrm { e } } } { \mathrm { M } _ { \mathrm { s } } } \right) \left( \frac { \mathrm { R } _ { \mathrm { e } } } { \mathrm { R } _ { \mathrm { m } } } \right) ^ { 3 }$
$\therefore\frac { \mathrm { M } _ { \mathrm { s } } } { \mathrm { M } _ { \mathrm { e } } } = \left( \frac { \mathrm { T } _ { \mathrm { m } } } { \mathrm { T } _ { \mathrm { e } } } \right) ^ { 2 } \left( \frac { \mathrm { R } _ { \mathrm { e } } } { \mathrm { R } _ { \mathrm { m } } } \right) ^ { 3 } = \frac { ( 393 ) ^ { 3 } } { 13 ^ { 2 } } = 3.56 \times 10 ^ { 5 }$

Q3.

Two planets of radii in the ratio 2 : 3 are made from the materials of density in the ratio 3 : 2. Then the ratio of acceleration due to gravity $g_1/g_2$ at the surface of two planets will be

1

B

2.5

C

$4 / 9$

D

0.12

##### Explanation

According to law of conservation of angular momentum, mvr = constant
$\Rightarrow \mathrm { vr } =$ constant
$\mathrm { v } _ { \max } \cdot \mathrm { r } _ { \min } = \mathrm { v } _ { \min } \cdot \mathrm { r } _ { \max }$
$\Rightarrow \frac { \mathrm { V } _ { \mathrm { B } } } { \mathrm { V } _ { \mathrm { A } } } = \frac { \mathrm { v } _ { \max } } { \mathrm { v } _ { \min } } = \frac { \mathrm { r } _ { \max } } { \mathrm { r } _ { \min } } = \mathrm { x }$

Q4.

A satellite of mass $\mathrm { m }$ is revolving in a circular orbit of radiusr. The relation between the angular momentum J of satellite and mass $\mathrm { m }$ of earth will be -

$\mathrm { J } = \sqrt { \mathrm { GMm } ^ { 2 } \mathrm { r } }$

B

$\mathrm { J } = \sqrt { \mathrm { GMm } }$

C

$\mathrm { J } = \sqrt { \mathrm { GMmr } }$

D

$\mathrm { J } = \sqrt { \frac { \mathrm { mr } } { \mathrm { M } } }$

Angular momentum of satellite, $\mathrm { J } =$ mvr. But,$\frac { \mathrm { GMm } } { \mathrm { r } ^ { 2 } } = \frac { \mathrm { mv } ^ { 2 } } { \mathrm { r } } \Rightarrow \mathrm { v } = \sqrt { \frac { \mathrm { GM } } { \mathrm { r } } } $$\therefore \mathrm { J } = \mathrm { m } \sqrt { \mathrm { GMr } } Correctly answered4 Incorrectly answered-1 Q5. A spaceship is launched into a circular orbit close to earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Radius of earth = 6400 \mathrm { km } , \mathrm { g } = 9.8 \mathrm { m } / \mathrm { sec } ^ { 2 } ) 3.285 \mathrm { km } / \mathrm { sec } B 32.85 \mathrm { m } / \mathrm { sec } C 11.32 \mathrm { km } / \mathrm { sec } D 7.32 \mathrm { m } / \mathrm { sec } ##### Explanation The orbital velocity of space ship, \mathrm { v } _ { 0 } = \sqrt { \frac { \mathrm { GM } } { \mathrm { r } } } If space, ship is very near to earth's surface, \mathrm { r } = Radius of earth = \mathrm { R } \quad \therefore \mathrm { v } _ { 0 } = \sqrt { \frac { \mathrm { GM } } { \mathrm { R } } } = \sqrt { \mathrm { Rg } } = \sqrt { 6.4 \times 10 ^ { 6 } \times 9.8 } = 7.995 \times 10 ^ { 3 } \mathrm { m } / \mathrm { sec } = 7.195 \mathrm { km } / \mathrm { sec } The escape velocity of space-ship \mathrm { v } _ { \mathrm { e } } = \sqrt { 2 \mathrm { Rg } } = 7.9195 \sqrt { 2 } = 11.2 \mathrm { km } / \mathrm { sec } Additional velocity required = 11.2 - 7.9195=3 .2805\,\, \mathrm { km /sec} Correctly answered4 Incorrectly answered-1 Q6. The ratio of the radius of the Earth to that of the moon is 10 . The ratio of g on earth to the moon is 6 . The ratio ofthe escape velocity from the earth's surface to that fromthe moon is approximately- A 10 8 C 4 D 2 ##### Explanation The escape velocity \mathrm { v } _ { \mathrm { e } } = \sqrt { 2 \mathrm { gR } } Now, \left( \mathrm { V } _ { \mathrm { e } } \right) _ { \text { moon } } = \sqrt { 2 \mathrm { gR } }$$ \left( \mathrm { V } _ { \mathrm { e } } \right) _ { \text { earth } } = \sqrt { 2 \times 6 \mathrm { g } \times 10 \mathrm { R } }$So $\frac { \left( \mathrm { V } _ { \mathrm { e } } \right) _ { \text { earth } } } { \left( \mathrm { V } _ { \mathrm { e } } \right) _ { \text { moon } } } = 8$

Q7.

Acceleration due to gravity on a planet is 10 times the valueon the earth. Escape velocity for the planet and the earthare $\mathrm { V } _ { \mathrm { p } }$ and $\mathrm { V } _ { \mathrm { e } }$ respectively. Assuming that the radii of theplanet and the earth are the same, then -

A

$\mathrm { V } _ { \mathrm { P } } = 10 \mathrm { V } _ { \mathrm { s } }$

$\mathrm { V } _ { \mathrm { P } } = \sqrt { 10 } \mathrm { V } _ { \mathrm { e } }$

C

$V _ { p } = \frac { V _ { e } } { \sqrt { 10 } }$

D

$V _ { P } = \frac { V _ { e } } { 10 }$

##### Explanation

Escape velocity $= \sqrt { \frac { 2 \mathrm { GM } } { \mathrm { R } } } = \sqrt { 2 \mathrm { gR } }$
$\therefore \frac { \mathrm { V } _ { \mathrm { p } } } { \mathrm { V } _ { \mathrm { e } } } = \sqrt { \frac { \mathrm { g } _ { \mathrm { p } } } { \mathrm { g } _ { \mathrm { e } } } \times \frac { \mathrm { R } _ { \mathrm { e } } } { \mathrm { R } _ { \mathrm { p } } } } = \sqrt { 10 \times 1 } = \sqrt { 10 }$
$\mathrm { V } _ { \mathrm { p } } = \sqrt { 10 } \mathrm { V } _ { \mathrm { e } }$
$\therefore \frac { \mathrm { V } _ { \mathrm { p } } } { \mathrm { V } _ { \mathrm { e } } } = \sqrt { \frac { \mathrm { g } _ { \mathrm { p } } } { \mathrm { g } _ { \mathrm { e } } } \times \frac { \mathrm { R } _ { \mathrm { e } } } { \mathrm { R } _ { \mathrm { p } } } } = \sqrt { 10 \times 1 } = \sqrt { 10 }$
$\mathrm { V } _ { \mathrm { p } } = \sqrt { 10 } \mathrm { V } _ { \mathrm { e } }$

Q8.

The Jupiter's period of revolution round the Sun is 12 timesthat of the Earth. Assuming the planetary orbits are circular,how many times the distance between the Jupiter and Sunexceeds that between the Earth and the sun.

5.242

B

4.242

C

3.242

D

2.242

##### Explanation

We know that $\mathrm { T } ^ { 2 } \propto \mathrm { a } ^ { 3 }$
Given that $( 12 \mathrm { T } ) ^ { 2 } \propto \mathrm { a } _ { 1 } ^ { 3 }$ and $\mathrm { T } ^ { 2 } \propto \mathrm { a } _ { 2 } ^ { 3 }$
$\therefore \frac { \mathrm { a } _ { 1 } ^ { 3 } } { \mathrm { a } _ { 2 } ^ { 3 } } = \frac { ( 12 \mathrm { T } ) ^ { 2 } } { \mathrm { T } ^ { 2 } } = 144$
or $\frac { \mathrm { a } _ { 1 } } { \mathrm { a } _ { 2 } } = ( 144 ) ^ { 1 / 3 } = 5.242$
Hence the jupiter's distance is 5.242 times that of theearth from the sun.

Q9.

The mean distance of mars from sun is 1.524 times the distance of the earth from the sun. The period of revolution of mars around sun will be-

2.88 earth year

B

1.88 earth year

C

3.88 earth year

D

4.88 earth year

##### Explanation

We know that $\mathrm { T } ^ { 2 } \propto \mathrm { a } ^ { 3 } \Rightarrow \mathrm { T } \propto ( \mathrm { a } ) ^ { 3 / 2 }$
$\therefore \frac { \mathrm { T } _ { \text { mars } } } { \mathrm { T } _ { \text { earth } } } = \left( \frac { \mathrm { a } _ { \text { mars } } } { \mathrm { a } _ { \text { earth } } } \right) ^ { \frac { 3 } { 2 } } = ( 1.524 ) ^ { 3 / 2 } = 1.88$
As the earth revolves round the sun in one year andhence, $\mathrm { T } _ { \text { earth } } = 1$ year.
$\therefore \mathrm { T } _ { \text { mars } } = \mathrm { T } _ { \text { earth } } \times 1.88 = 1 \times 1.88 = 1.88$ earth-year.

Q10.

The semi-major axes of the orbits of mercury and mars are respectively 0.387 and 1.524 in astronomical unit. If the period of Mercury is 0.241 year, what is the period of Mars.

A

1.2 years

B

3.2 years

C

3.9 years

1.9 years

##### Explanation

$\frac{\mathrm{T}_{\text { mercury }}}{\mathrm{T}_{\text { mars }}}=\left(\frac{\mathrm{a}_{\text { mercury }}}{\mathrm{a}_{\text { mars }}}\right)^{3 / 2}=\left(\frac{0.387}{1.524}\right)^{3 / 2}$$\therefore \mathrm{T}_{\mathrm{mars}}=\mathrm{T}_{\mathrm{mercury}} \times\left(\frac{1.524}{0.387}\right)^{3 / 2}$$=(0.241 \text { years) } \times(7.8)=1.9 \text { years. }$

Q1.

Two radioactive samples have activity ratio $\mathrm {5 : 1}.$ If the temperature of samples is made 2 times, then the activity ratio is

A

$\mathrm {1 : 5}$

$5 : 1$

C

$\mathrm {1 : 2.5}$

D

$2.5 : 1$

##### Explanation

It is independent of the temperature.

Q2.

The velocity of centre of mass of sphere in previous question when pure rolling starts will be equal to

A

$\frac { 2 \mathrm { V } _ { 0 } } { 5 }$

B

$\frac { 5 \mathrm { V } _ { 0 } } { 6 }$

$\frac { V _ { 0 } } { 7 }$

D

0

##### Explanation

COAM: about point of contact $\mathrm { Mv } _ { 0 } \mathrm { R } - \frac { 2 } { 5 } \mathrm { MR } ^ { 2 } \left( \frac { 2 \mathrm { v } _ { 0 } } { \mathrm { R } } \right) = \frac { 7 } { 5 } \mathrm { MvR } ; \mathrm { v } = \frac { \mathrm { V } _ { 0 } } { 7 }$

Q3.

Angle between current through inductor and capacitor will be

$143 ^ { \circ }$

B

$90 ^ { \circ }$

C

$53 ^ { \circ }$

D

None

##### Explanation

$i _ { \mathrm { L } } = 40 \sqrt { 2 } \sin \left( \omega \mathrm { t } + \frac { \pi } { 4 } - 53 ^ { \circ } \right) $$i _ { \mathrm { C } } = 20 \sqrt { 2 } . \sin \left( \omega \mathrm { t } + \frac { \pi } { 4 } + 90 ^ { \circ } \right)$$ \therefore \quad$ angle between $i _ { \mathrm { L } }$ and $i _ { \mathrm { C } }$ is $\left( 90 + 53 ^ { \circ } \right) = 143 ^ { \circ }$

Q4.

Potential drop across $\mathrm { X } _ { \mathrm { L } }$ by a. . voltmeter will be

160 volt

B

120 volt

C

200 volt

D

$160\sqrt { 2 } \mathrm { volt }$

$\mathrm { i } _ { \mathrm { ms } } ($ through inductor $) = 40 \mathrm { A } $$\therefore \quad \mathrm { V } \left( Potential drop across \mathrm { X } _ { \mathrm { L } } \right)$$ \quad = 40 \times 4 = 160 \mathrm { Volt } $$\therefore Correct answer ( \mathrm { A } ) Correctly answered4 Incorrectly answered-1 Q5. When a dilute, aqueous solution of potassium permanganate is run from a burette into a flask. Contain in dilute, aqueous oxalic acid and dilute sulphuric acid, the rate of reaction suddenly increases considerably as more potassium permaganate is added. The reason for this is that the manganese (II) ions produced catalyse the reaction B the \mathrm { pH } of solution in the flask increases. C the reaction is exothermic and the heat energy liberated affects the rate. D the sulphuric acid removed water and so causes the reaction to proceed more rapidly to completion. ##### Explanation In acidic medium, MnO _ { 4 } reduces into Mn(ll), which catalyses the reaction. The catalytic action of Mn(ll)may be represented as Correctly answered4 Incorrectly answered-1 Q6. For the same reaction between \mathrm {V}_ { \infty } , \mathrm {V} _ { 0 } and \mathrm {{V}_ { t }} after 266.4 minutes will be A 99 \mathrm { V } _ { \mathrm { t } } = 98 \mathrm { V } _ { \infty } + \mathrm { V } _ { 0 } 99 \mathrm { V } _ { \infty } = 98 \mathrm { V } _ { \mathrm { t } } + \mathrm { V } _ { 0 } C 100 \mathrm { V } _ { 0 } = 99 \mathrm { V } _ { \infty } - \mathrm { V } _ { \mathrm { t } } D 100 \mathrm { V } _ { \mathrm { t } } = 99 \mathrm { V } _ { \infty } - \mathrm { V } _ { 0 } ##### Explanation \begin{array} { l } { \mathrm { n } = \frac { t } { t _ { 1 / 2 } } = \frac { 266.4 } { 40 } = 6.66 } \\ { \mathrm { V } _ { \mathrm { m } } - \mathrm { V } _ { 0 } = 99 \mathrm { V } _ { \infty } - 99 \mathrm { V } _ { \mathrm { t } } } \\ { 99 \mathrm { V } _ { \mathrm { t } } = 98 \mathrm { V } _ { \infty } + \mathrm { V } _ { 0 } } \end{array} \quad \Rightarrow \frac { V _ { \infty } - V _ { 0 } } { V _ { \infty } - V _ { t } } = 99 Correctly answered4 Incorrectly answered-1 Q7. A hollow closed conductor of irregular shape is given some charge. Which of the following statements are incorrect? A The entire charge will appear on its outer surface B All points on the conductor will have same potential All points on its surface will have the same charge density All points near its surface (just inside) and outside it will have the same electric field intensity ##### Explanation ( \mathrm { A } , \mathrm { B } ) Correctly answered4 Incorrectly answered-1 Q8. Backlash error may occur in which of the following instruments? A Slide calipers and screw gauge Spherometer and screw gauge C Slide cappliers and spherometer D Slide calipers, screw gauge and spherometer ##### Explanation Backlash error occurs only in instruments using screws Correctly answered4 Incorrectly answered-1 Q9. In a resonance-column experiment to measure the velocity of sound, the first resonance isobtained at a length 1 _ { 1 } and the second resonance at a length l _ { 2 } . Then which of the following is incorrect l _ { 2 } > 3 l _ { 1 } l _ { 2 } = 3 l _ { 1 } C l _ { 2 } < 3 l _ { 1 } May be any of the above, depending on the frequency of the tuning fork used ##### Explanation Sol. \lambda \rightarrow wavelength \delta \rightarrow end correction l _ { 1 } = \frac { \lambda } { 4 } + \delta ; l _ { 2 } = \frac { 3 \lambda } { 4 } + \delta or l _ { 2 } = 3 \left( l _ { 1 } - \delta \right) + \delta$$ = 3 l _ { 1 } - 2 \mathrm { d }$

Q10.

In experiments using the metre bridge, post-office box, and potentiometer, a galvanometer is used. Which property of the galvanometer makes it suitable for these experiments?

A

It has a relatively high coil resistance

B

It measures the magnitude of the current

It can indicate currents flowing through it in either direction

It is very sensitive to small currents

##### Explanation

In the three experiments, the jockey is moved according to the direction of deflection of the galvanometer.

Q1.

The mechanism of drug resistance in MDR Tb is

A

Transformation

Transduction

C

Conjugation

D

Mutation

##### Explanation

Mutation in the chromosome

Q2.

Phage typing is useful epidemiological tool in all except

A

Salmonella

B

Staph aureus

Vibrio cholerae

D

Shigella

##### Explanation

The phage typing in Salmonella is carried out based on the Vi antigen, hence the non Vi Salmonella cannot be typed and so the differentiation is limited

Q3.

A disinfectant was used to reduce the number of bacteria present in the biosafety cabinet. Assuming that there were 103 bacteria initially, how many would be left after 20 minutes if 90% killing takes place in 10 minutes

A

100

B

10

C

1

0

##### Explanation

In the first 10 minutes the count comes down to 100 and in the subsequent 10 min the count comes to 10

Q4.

The C. tetani smooth strain for vaccine production should be grown in liquid media for how much time to extract the toxin by ultrafiltration

A

7 days

B

48 hours

6 hours

D

4 hours

##### Explanation

7 days for the lytic cycle to occur

Q5.

Allotype of Immunoglobulins are due to

A

Alternative RNA splicing

B

Gene rearrangement

C

Recombination of VD-J genes

Allelic exclusion

##### Explanation

Allotypes i.e. subtle amino acid differences in the constant region of Ig among different individuals within a species is due to allelic exclusion

Q6.

All are true about ?2 microglobulin deficiency except

A

Mutation at 283 position of hfe gene

Cysteine is replaced by tyrosine

C

No inhibition of dietary iron absorption

D

Defect at HLA ?1 domain

##### Explanation

Defect at HLA ?3 domain

Q7.

A boy presents with lymphocytosis and hypocellularity in both cortical and paracortical areas of the lymph node. What is the deficient leukocyte surface molecule

Integrins

B

L-selectins

C

D

##### Explanation

L-selectins are the adhesion molecules on lymphocytes for binding to the high endothelial cells

Q8.

Which factor introduced with a vaccine would increase the immunogenicity of the vaccine

IgG

B

Complement component C3a

C

IL-8

D

TNF?

##### Explanation

IL8 being a chemotactic factor for the neutrophils

Q9.

The immunoglobulin which shows highest affinity is

IgG

B

IgM

C

IgD

D

IgA

##### Explanation

IgG has a prolonged response so the B cells get the opportunity to carry out mutation and repeated clonal selection and proliferation to increase the affinity

Q10.

The predominant B cell stage seen in Bruton's X linked hypergammaglobulinemia is

Pro B cell

B

Pre B cell

C

Mature B cells

D

Memory B cells

##### Explanation

Due to deficiency of tyrosine kinase the pre B cells cannot proliferate to mature cells.

Q1.

The LCM of two numbers is 864 and their HCF is $144 .$ If one of the number is $288 ,$ the other number is :

A

576

B

1296

432

D

144

##### Explanation

Using Rule 1 Required number$= \frac { \mathrm { LCM } \times \mathrm { HCF } } { \text { First number } }$ $= \frac { 864 \times 144 } { 288 } = 432$

Q2.

LCM of two numbers is 225 and their HCF is $5 .$ If one number is $25 ,$ the other number will be:

A

5

B

25

45

D

225

##### Explanation

Using Rule 1,LCM x HCF= 1st Number 2nd Number $\Rightarrow 225 \times 5=25 \times x$âˆ´ X=$\frac{225 \times 5}{25}=45$

Q3.

The L.C.M. of two numbers is 1820 and their H.C.F. is 26. If one number is 130 then the other number is:

A

70

B

1690

364

D

1264

##### Explanation

Using Rule 1,Given thatL.C.M. of two numbers $= 1820$H.C.F. of two numbers $= 26$One of the number is 130$\therefore$ Another number$= \frac { 1820 \times 26 } { 130 } = 364$

Q4.

The LCM of two numbers is1920 and their HCF is $16 .$ If oneof the number is $128 ,$ find theother number.

A

204

240

C

260

D

320

##### Explanation

Using Rule 1 We have,First number $\times$ second number$= \mathrm { LCM } \times \mathrm { HCF }$ $\therefore$ Second number $= \frac { 1920 \times 16 } { 128 } = 240$

Q5.

The HCF of two numbers 12906 and 14818 is $478 .$ Their $L C M$ is

400086

B

200043

C

600129

D

800172

##### Explanation

Using Rule 1, Product of two numbers $=$ HCF $\times \mathrm { LCM }$ $\Rightarrow 12906 \times 14818$ $= \mathrm { LCM } \times 478$ $\Rightarrow \mathrm { LCM } = \frac { 12906 \times 14818 } { 478 }$ $= 400086$

Q6.

The H.C.F. and L.C.M. of two $2 -$ digit numbers are 16 and 480 respectively. The numbers are:

A

$40,48$

B

$60,72$

C

$64,80$

$80,96$

##### Explanation

Using Rule 1. H.C.F of the two 2-digit numbers= 16 Hence, the numbers can be expressed as 16x and 16 y, where x and y are prime to each other. Now First number x second number = H.C.F x L.C.M.
$\Rightarrow$ 16x x 16y=16 x 480
$\Rightarrow$ xy=$\frac{16 \times 480}{16 \times 16}=30$The possible pairs of x and y, setisfying the condition xy=30 are:
(3,10),(5,6)(1,30),(2,15)
Since the numbers are of $2 -$ digits each. Hence, admissible pair is $( 5,6 )$ $\therefore$ Numbers are : $16 \times 5 = 80$ and $16 \times 6 = 96$

Q7.

The HCF of two numbers is 16 and their LCM is $160 .$ If one of the number is $32 ,$ then the other number is

A

48

80

C

96

D

112

##### Explanation

Using Rule 1 We know that, First number $\times$ Second number $= \mathrm { LCM } \times \mathrm { HCF }$ $\Rightarrow$ Second number $= \frac { 16 \times 160 } { 32 } = 80$

Q1.

The product of two numbers is$4107 .$ If the H.C.F. of the numbers is 37 , the greater number is

A

185

111

C

107

D

101

##### Explanation

$\mathrm { Using \, Rule \, 1 ,\\ LCM= \frac {Product \, of \, two\, numbers}{HCF}\\=\frac {4107}{37}=111\\ Obviously,\,numbers\, are \, 111 \, and \\ 37 \, which \, satisfy\, the\, given\, condition.\\Hence,\, the \, greater\, number \,=111}$

Q9.

The HCF of two numbers is 15 and their $L C M$ is $300 .$ If one of the number is 60 , the other is:

50

75

C

65

D

100

##### Explanation

Using Rule 1 First number $\times$ Second number $= \mathrm {HCF}$ $\times \mathrm { LCM }$ $\therefore$ Second number $= \frac { 15 \times 300 } { 60 } = 75$

Q10.

The HCF and LCM of two numbers are 12 and 924 respectively.Then the number of such pairs is

A

0

B

1

2

D

3

##### Explanation

Let the numbers be 12$x$ and 12$y$ where $x$ and $y$ are prime to each other. $\therefore \mathrm { LCM } = 12 x y$ $\therefore 12 x y = 924$ $\Rightarrow x y = 77$ $\therefore$ Possible pairs $= ( 1,77 )$ and $( 7,11 )$

Q1.

The percentage of pyridine $(C_5H_5N)$ that forms pyridinium ion $(C_5H_5N^+H)$ in a 0.10 M aqueous pyridine solution $\mathrm{(K_b\,\,for\,\,C_5H_5N=1.7 \times 10^{-9})}$ is

A

0.0060$\%$

0.013$\%$

C

0.77$\%$

D

1.6$\%$

##### Explanation

$\mathrm { C } _ { 5 } \mathrm { H } _ { 5 } \mathrm { N } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { C } _ { 5 } \mathrm { H } _ { 5 } ^ { + } \mathrm { H } + \mathrm { OH } ^ { - }$0.10$\mathrm { M }$
$\alpha = \sqrt { \frac { K _ { b } } { C } } = \sqrt { \frac { 1.7 \times 10 ^ { - 9 } } { 0.10 } } = 1.30 \times 10 ^ { - 4 }$
$\therefore$ Percentage of pyridine that forms pyridiniumion $= 1.30 \times 10 ^ { - 4 } \times 100 = 0.013 \%$

Q2.

Which of the following fluoro-compounds ismost likely to behave as a Lewis base?

A

$\mathrm { BF } _ { 3 }$

$\mathrm { PF } _ { 3 }$

C

$\mathrm { CF } _ { 4 }$

D

$\mathrm { SiF } _ { 4 }$

$\mathrm { BF } _ { 3 } \rightarrow$ Lewis acid (incomplete octet)$\mathrm { PF } _ { 3 } \rightarrow$ Lewis base (presence of lone pair on $\mathrm { P }$ atom $) $$\mathrm { CF } _ { 4 } \rightarrow Complete octet \mathrm { SiF } _ { 4 } \rightarrow Lewis acid (empty d -orbital in Si-atom) Correctly answered4 Incorrectly answered-1 Q3. M Y and N Y _ { 3 } , two nearly insoluble salts, have the same K _ { s p } values of 6.2 \times 10 ^ { - 13 } at room temperature. Which statement would be truein regard to M Y and N Y _ { 3 } ? A The salts M Y and N Y _ { 3 } are more soluble in 0.5 \mathrm { M } K Y than in pure water. B The addition of the salt of K Y to solutionof M Y and N Y _ { 3 } will have no effect on their solubilities. C The molar solubilities of M Y and N Y _ { 3 } in water are identical. The molar solubility of M Y in water is less than that of N Y _ { 3 } . ##### Explanation \quad ( \mathrm { d } ) : For \mathrm M\mathrm Y :\mathrm K _ { s p } = s _ { 1 } ^ { 2 }$$ \Rightarrow s _ { 1 } = \sqrt {\mathrm K _ { s p } } = \sqrt { 6.2 \times 10 ^ { - 13 } } = 7.87 \times 10 ^ { - 7 } \mathrm { mol } \mathrm { L } ^ { - 1 }$For $\mathrm N\mathrm Y _ { 3 } :\mathrm K _ { s p } = 27 s _ { 2 } ^ { 4 } $$\Rightarrow \mathrm { s } _ { 2 } = \sqrt [ 4 ] { \frac { 6.2 \times 10 ^ { - 13 } } { 27 } } = 3.89 \times 10 ^ { - 4 } \mathrm { mol } \mathrm { L } ^ { - 1 } Hence, molar solubility of \mathrm M\mathrm Y in water is less thanthat of \mathrm N\mathrm Y _ { 3 } . Correctly answered4 Incorrectly answered-1 Q4. If the equilibrium constant for \mathrm { N } _ { 2 ( g ) } + \mathrm { O } _ { 2 ( g ) } \rightleftharpoons 2 \mathrm { NO } _ { ( g ) } is K , the equilibriumconstant for \frac { 1 } { 2 } \mathrm { N } _ { 2 ( g ) } + \frac { 1 } { 2 } \mathrm { O } _ { 2 ( g ) } \rightleftharpoons \mathrm { NO } _ { ( g ) } will be A \frac { 1 } { 2 } K B K C K ^ { 2 } K ^ { 1 / 2 } ##### Explanation If the reaction is multiplied by \frac { 1 } { 2 } , then new equilibrium constant, K ^ { \prime } = K ^ { 1 / 2 } Correctly answered4 Incorrectly answered-1 Q5. What is the pH of the resulting solution when equal volumes of 0.1 \mathrm { M } \mathrm { NaOH } and 0.01 \mathrm { MHCl } are mixed? A 2 B 7 C 1.04 12.65 ##### Explanation One mole of NaOH is completely neutralised by one mole of HCl.Hence, 0.01 mole of NaOH will be completely neutralised by 0.01 mole of HCl. \Rightarrow NaOH left unneutralised =0.1-0.01=0.09 molAs equal volumes of two solutions are mixed, [ \mathrm { OH } ] ^ { - } = \frac { 0.09 } { 2 } = 0.045 \mathrm { M } \Rightarrow \mathrm { pOH } = - \log ( 0.045 ) = 1.35 \therefore \quad \mathrm { pH } = 14 - 1.35 = 12.65 Correctly answered4 Incorrectly answered-1 Q6. Aqueous solution of which of the following compounds is the best conductor of electric current? Hydrochloric acid, HCl B Ammonia, NH_{3} C Fructose, \mathrm { C } _ { 6 } \mathrm { H } _ { 12 } \mathrm { O } _ { 6 } D Acetic acid, \mathrm { C } _ { 2 } \mathrm { H } _ { 4 } \mathrm { O } _ { 2 } ##### Explanation HCl is a strong acid and dissociates completely into ions in aqueous solution. Correctly answered4 Incorrectly answered-1 Q7. Which one of the following pairs of solutionis not an acidic buffer? A \mathrm { CH } _ { 3 } \mathrm { COOH } and \mathrm { CH } _ { 3 } \mathrm { COONa } B \mathrm { H } _ { 2 } \mathrm { CO } _ { 3 } and \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } C \mathrm { H } _ { 3 } \mathrm { PO } _ { 4 } and \mathrm { Na } _ { 3 } \mathrm { PO } _ { 4 } \mathrm { HClO } _ { 4 } and \mathrm { NaClO } _ { 4 } ##### Explanation Acidic buffer is a mixture of a weak acid andits salt with a strong base. HClO_4 is a strong acid. Correctly answered4 Incorrectly answered-1 Q8. Which of the following statements is correctfor a reversible process in a state of equilibrium? \Delta G ^ { \circ } = - 2.30 \,\,R T\,\, \log K B \Delta G ^ { \circ } = 2.30 R T \log K C \Delta G = - 2.30 R T \log K D \Delta G = 2.30 \,\,R T\,\, \log\,\, K ##### Explanation ( \mathbf { a } ) Correctly answered4 Incorrectly answered-1 Q9. If the value of equilibrium constant for aparticular reaction is 1.6 \times 10 ^ { 12 } , then at equilibrium the system will contain mostly products B similar amounts of reactants and products C all reactants D mostly reactants. ##### Explanation The value of K is high which means reaction proceeds almost to completion i.e., the system will contain mostly products. Correctly answered4 Incorrectly answered-1 Q10. Which of the following salts will give highest \mathrm { pH } in water? A \mathrm { KCl } B \mathrm { NaCl } \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } D \mathrm { CuSO } _ { 4 } ##### Explanation \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } which is a salt of \mathrm{NaOH} (strongbase) and \mathrm { H } _ { 2 } \mathrm { CO } _ { 3 } (weak acid) will produce a basic solution with pH greater than 7 . Correct answered1 Incorrect answered -0.25 Q1. 12 men take 18 days to complete a job whereas 12 women in18 days can complete \frac { 3 } { 4 } of the same job. How many dayswill 10 men and 8 of the same take to complete thesame job? A 6 13 \frac { 1 } { 2 } C 12 D Data inadequate ##### Explanation 12 M \times 18 = 12 W \times 18 \times \frac { 4 } { 3 }$$ \therefore \quad 1 W = \frac { 3 } { 4 } M $$10 M + 8 W = 10 M + 8 \times \frac { 3 } { 4 } M = 16 M$$ \therefore \quad 16$ men can complete the same workin $\frac { 12 \times 18 } { 16 } = \frac { 27 } { 2 } = 13 \frac { 1 } { 2 }$ days

Q2.

The work done by a woman in 8 hours is equal to the workdone by a man in 6 hours and by a boy in 12 hours. Ifworking 6 hours per day, 9 men can complete a work in 6days then in how many days can 12 men, 12 women and 12boys together finish the same work by working 8 hours perday?

A

1$\frac { 1 } { 3 } \mathrm { days }$

B

3$\frac { 2 } { 3 }$ days

C

3 days

1$\frac { 1 } { 2 } \mathrm { days }$

##### Explanation

$\begin{array} { l } { \text { } 8 W = 6 M = 12 B } \\ { 12 M + 12 W + 12 B \Rightarrow 12 M + 9 M + 6 M = 27 M } \\ { \therefore } & { 9 \text { men can complete the work by working } 1 \text { hour per } } \\ { \text { day in } 6 \times 6 \text { days } } \\ { \therefore \quad 27\text { men working } 8 \text { hours per day } = \frac { 6 \times 6 \times 9 } { 27 \times 8 } = 1 \frac { 1 } { 2 } \text { days. } } \end{array}$

Q3.

Suresh can complete a job in 15 hours. Ashutosh alone cancomplete the same job in 10 hours. Suresh works for 9 hoursand then the remaining job is completed by Ashutosh. Howmany hours will it take Ashutosh to complete the remainingjob alone?

4

B

5

C

6

D

12

##### Explanation

The part of job that Suresh completes in 9 hours $= \frac { 9 } { 15 } = \frac { 3 } { 5 }$Remaining job $= 1 - \frac { 3 } { 5 } = \frac { 2 } { 5 }$Remaining job can be done by Ashutosh in $\frac { 2 } { 5 } \times 10 = 4$ hours

Q4.

10 men and 15 women finish a work in 6 days. One mafealone finishes that work in 100 days. In how many days will $C$a woman finish the work?

A

125 days

B

150 days

C

245 days

225 days

##### Explanation

\begin{aligned} \text { } & 15 \text { women's work of a day } = \frac { 1 } { 6 } - \frac { 1 } { 10 } \Rightarrow \frac { 1 } { 15 } \text { part } \\ & \therefore \text { for } 1 \text { whole part a woman will take } \\ & = 15 \times 15 = 225 \text { days. } \end{aligned}

Q5.

24 men working 8 hours a day can finish a work in 10 days.Working at the rate of 10 hours a day, the number of menrequired to finish the same work in 6 days is:

A

30

32

C

34

D

36

$\quad m _ { 1 } \times d _ { 1 } \times t _ { 1 } \times w _ { 2 } = m _ { 2 } \times d _ { 2 } \times t _ { 2 } \times w _ { 1 } $$24 \times 10 \times 8 \times 1 = m _ { 2 } \times 6 \times 10 \times 1 \Rightarrow m _ { 2 } = \frac { 24 \times 10 \times 8 } { 6 \times 10 } = 32 \mathrm { men } Correctly answered1 Incorrectly answered-0.25 Q6. A water tank has three taps A, Band C. Tap A, when opened,can fill the water tank alone in 4 hours. Tap B, when opened,can fill the water tank alone in 6 hours and tap C, whenopened, can empty the water tank alone in 3 hours. If tapsA, Band C are opened simultaneously, how long will it taketo fill the tank completely? A 10 hours B 8 hours C 18 hours 12 hours ##### Explanation Required time to fill the tank = \frac { 1 } { \left( \frac { 1 } { 4 } + \frac { 1 } { 6 } \right) - \frac { 1 } { 3 } } = \frac { 1 } { \frac { 5 } { 12 } - \frac { 1 } { 3 } } = \frac { 1 } { \frac { 1 } { 12 } } = 12 \mathrm { h } Correctly answered1 Incorrectly answered-0.25 Q7. 10 men can complete a piece of work in 15 days and 15women can complete the same work in 12 days. If all the 10men and 15 women work together, in how many days willthe work get completed? A 6 B 7 \frac { 2 } { 3 } 6 \frac { 2 } { 3 } D 6 \frac { 1 } { 3 } ##### Explanation 10 \mathrm { men } + 15 women in 1 day do \frac { 1 } { 15 } + \frac { 1 } { 12 } = \frac { 9 } { 60 } work \therefore Time taken = \frac { 60 } { 9 } \mathrm { days } = 6 \frac { 2 } { 3 } \mathrm { days } Correctly answered1 Incorrectly answered-0.25 Q1. Rajani has to read a book of 445 pages. She has alreadyread the first 157 pages of the book and if she reads 24pages of the book everyday then how long will she takenow to complete the book? A 25 days B 20 days C 46 days D 21 days None of these ##### Explanation Remaining pages to read = 445 - 157 = 288$$ \therefore$ Reqd. number of days $= \frac { 288 } { 24 } = 12$

Q9.

10 horses and 15 cows eat grass of 5 acres in a certain time.How many acres will feed 15 horses and 10 cows for thesame time, supposing a horse eats as much as 2 cows?

40$/ 7$ acres

B

39$/ 8$ acres

C

40$/ 11$ acres

D

25$/ 9$ acres

##### Explanation

$\begin{array} { l l } { \text { } } & { 1 \text { horse } = 2 \text { cows, } 10 \text { horses } = 20 \text { cows. } } \\ { } & { \Rightarrow 10 \text { horses } + 15 \text { cows } = 20 + 15 = 35 \text { cows. } } \\ { } & { 15 \text { horses } + 10 \text { cows } = 40 \text { cows. Now } 35 \text { cows eat } 5 } \\ { } & { \text { acres. } } \end{array}$
$\Rightarrow 40 \quad cows \quad eat \quad 5 \times \frac{40}{35}=5 \frac{5}{7}.$
Here we have converted everything in terms of cows, you can work in terms of horses also

Q10.

$\mathrm { X }$ and $\mathrm { Y }$ can do job in 25 days and 30 days respectively.They work together for 5 days and then $\mathrm { X }$ leaves. Y willfinish the rest of the work in how many days?

A

18 days

19$\mathrm { days }$

C

20$\mathrm { days }$

D

21 days

##### Explanation

X's one day's work=$\frac{1}{25}th$part of whole work
Y's one day's work =$\frac{1}{30}th$ part of whole work.
Their one day's work =$\frac{1}{25}+ \frac{1}{30}= \frac{1}{150}$ part of whole work.
Now, work is done in 5 days $= \frac { 11 } { 150 } \times 5 = \frac { 11 } { 30 }$ th
$\therefore$ Remaining work $= 1 - \frac { 11 } { 30 } = \frac { 19 } { 30 }$ th of whole work
Now, $\frac { 1 } { 30 }$ th work is done by $Y$ in one day.
$\therefore \frac { 19 } { 30 }$ th work is done by Y in $\frac { 1 } { 1 / 30 } \times \frac { 19 } { 30 } = 19$ days

Q1.

Vultures which used to be very common in Indian countryside some years ago are rarely seen nowadays. This is attributed to

A

the destruction of their nesting sites by new invasive species

a drug used by cattle owners for treating their diseased cattle

C

scarcity of food available to them

D

a widespread, persistent and fatal disease among them

##### Explanation

Biologists think that many vultures being poisoned by drug known as diclofenac as farmer administered to livestock. Such drug wrecks the kidneys of vultures that search the remains and kills the vulture in days, so this makes option B. correct.

Q2.

What would happen if phytoplankton of an ocean is completely destroyed for some reason?
1) The ocean as a carbon sink would be adversely affected.
2) The food chains in the ocean would be adversely affected.
3) The density of ocean water would drastically decrease.
Select the correct answer using the codes given below:

1 and 2 only

B

2 only

C

3 only

D

1, 2 and 3

##### Explanation

Statement 1 is correct as phytoplankton knock back carbon dioxide generating world's oxygen output. Statement 2 is correct as Phytoplankton serves as initial link in oceanic food chain by eating zooplankton which further eats other animals that are consumed by sea creatures, so option A. is correct as only statements 1 and 2 are correct.

Q3.

Consider the following agricultural practices:
1) Contour bunding
2) Relay cropping
3) Zero tillage
In the context of global climate change, which of the above helps/help in carbon sequestration/storage in the soil?

A

1 and 2 only

B

3 only

1, 2 and 3

D

None of them

##### Explanation

Carbon sequestration involves carbon capturing with long atmospheric carbon dioxide storage which defers global warming. Statement 1 is correct as contour bunding, agricultural practice is carried across hill slopes is an effective method of soil conservation. Statement 2 is correct as Relay cropping involves growing of two or more crops on similar field with planting of 2nd crop after 1st development. Statement 3 is correct as Zero Tillage is applied in conservation of agriculture which brings quantum leap in crop production.

Q4.

Consider the following:
1) Black-necked crane
2) Cheetah
3) Flying squirrel
4) Snow leopard
Which of the above are naturally found in India?

A

1, 2 and 3 only

1, 3 and 4 only

C

2 and 4 only

D

1, 2, 3 and 4

##### Explanation

Statement 1 is correct, black-necked crane is medium-sized crane in Asia which is breeds in remote parts of India and Bhutan. Statement 2 is incorrect as Cheetah like grasslands, savannahs, semi- desert areas and was initially found in Kavir desert of Iran. Statement 3 is correct as flying squirrel naturally found in China, India, Indonesia, Myanmar, Sri Lanka, Taiwan and Vietnam. Statement 4 is correct as snow leopard found on mountain ranges of Central and South Asia which serves as one of the five big cats found in India. We see that statements 1, 3 and 4 are correct, so best option is B.

Q5.

Which of the following can be threats to the biodiversity of a geographical area?
1) Global warming
2) Fragmentation of habitat
3) Invasion of alien species
4) Promotion of vegetarianism
Select the correct answer using the codes given below:

1, 2 and 3 only

B

2 and 3 only

C

1 and 4 only

D

1, 2, 3 and 4

##### Explanation

Statement 2 is correct biodiversity threat result with Loss of Habitat and fragmentation with deforestation in tropical forests result as cause of mass extinctions lead by human activity. Statement 1 is correct as biodiversity threat results could be Global warming which occurs due to changing global climate which lead to threatening of species and ecosystems. Statement 3 is correct as invasion of non-native species result in important and unseen reason of extinctions, so correct option is A, as three statements are correct.

Q6.

What is the difference between the antelopes Oryx and Chiru?

Oryx is adapted to live in hot and arid areas whereas Chiru is adapted to live in steppes and semi-desert areas of cold high mountains

B

Oryx is poached for its antlers whereas Chiru is poached for its musk

C

Oryx exists in western India only whereas Chiru exists in north-east India only

D

None of the statements (a), (b) and (c) given above is correct

##### Explanation

Oryx is tailored to survive in hot and arid areas while Chiru is tailored to exist in steppes and semi-desert areas of cold high mountains, which makes option A correct. It is seen that Oryx species exists near desert conditions which lives without water for longer time. Tibetan antelope or chiru lives in cold high mountains states option B and C incorrect.

Q7.

The Millennium Ecosystem Assessment describes the following major categories of ecosystem services-provisioning, supporting, regulating, preserving and cultural. Which one of the following is supporting service?

A

Production of food and water

B

Control of climate and disease

Nutrient cycling and crop pollination

D

Maintenance of diversity

##### Explanation

Millennium Ecosystem Assessment assesses the consequences of ecosystem change for human well-being describing conditions of ecosystems and consequences of change. It is noted that MEA involves in nutrients cycling along with pollination of different crops which states that option C. is correct.

Q1.

Consider the following statements:
If there were no phenomenon of capillarity
1) it would be difficult to use a kerosene lamp
2) one would not be able to use a straw to consume a soft drink
3) the blotting paper would fail to function
4) the big trees that we see around would not have grown on the Earth
Which of the statements given above are correct?

A

1, 2 and 3 only

1, 3 and 4 only

C

2 and 4 only

D

1, 2, 3 and 4

##### Explanation

Capillarity serves as capacity of flow of liquid in small space without gravity. Statement 1 is correct as kerosene lamp involves phenomena of capillary action. Statement 2 is incorrect as capillary action is not applied in case of application of straw to consume soft drink. Statement 3 is correct as blotting paper will not work without phenomena of capillary action. Statement 4 is correct as capillary action involves small tree to grow under big trees, so correct option is B.

Q9.

Which one of the following groups of animals belongs to the category of endangered species?

Great Indian Bustard, Musk Deer, Red Panda and Asiatic Wild Ass

B

Kashmir Stag, Cheetal, Blue Bull and Great Indian Bustard

C

Snow Leopard, Swamp Deer, Rhesus Monkey and Saras (Crane)

D

Lion-tailed Macaque, Blue Bull, Hanuman Langur and Cheetal

##### Explanation

Endangered species are risky which are threatened by factors like habitat loss, hunting, disease and climate change, hence we see that Great Indian Bustard, Musk Deer, Red Panda and Asiatic Wild Ass mentioned in option A. is correct.

Q10.

Consider the following kinds of organisms:
1) Bat
2) Bee
3) Bird
Which of the above is/are pollinating agent/agents?

A

1 and 2 only

B

2 only

C

1 and 3 only

1, 2 and 3

##### Explanation

Pollinator is an agent which moves pollen, whether it be bees, flies, bats, moths or birds, so 1, 2 and 3 are correct which makes option D. correct.