Explore popular questions from Physical World and Measurement for JEE Advanced. This collection covers Physical World and Measurement previous year JEE Advanced questions hand picked by experienced teachers.

## Chemistry

Physical World and Measurement

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Q 1. The dimension of {tex} \left(\frac { 1 } { 2 } \right) \varepsilon _ { 0 } E ^ { 2 }{/tex}({tex} \varepsilon _ { 0 }: {/tex} permittivity of free space, {tex} E{/tex} electric field)

A

{tex} M L T ^ { - 1 } {/tex}

B

{tex} M L ^ { 2 } T ^ { - 2 } {/tex}

{tex} M L ^ { - 1 } T ^ { - 2 } {/tex}

D

{tex} M L ^ { 2 } T ^ { - 1 } {/tex}

##### Explanation

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Q 2. A wire of length {tex} \ell = 6 \pm 0.06 \mathrm { cm } {/tex} and radius {tex} r = 0.5 \pm 0.005 \mathrm { cm } {/tex} and mass {tex} m = 0.3 \pm 0.003 \mathrm { gm } . {/tex} Maximum percentage error in density is

4

B

2

C

1

D

6.8

##### Explanation

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Q 3. In a screw gauge, the zero of mainscale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are {tex} 50 . {/tex} It moves {tex} 0.5 \mathrm { mm } {/tex} on main scale in one rotation. The diameter of the ball in figure (ii) is

A

{tex} 2.25 \mathrm { mm } {/tex}

B

{tex} 2.20 \mathrm { mm } {/tex}

{tex} 1.20 \mathrm { mm } {/tex}

D

{tex} 1.25 \mathrm { mm } {/tex}

##### Explanation

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Q 4. In the determination of Young's modulus {tex}\left(Y = \frac{4MLg}{\pi l d^2}\right){/tex} by using Searle's method, a wire of length {tex}L{/tex} = 2 m and diameter {tex}d{/tex} = 0.5 mm is used. For a load {tex}M{/tex} = 2.5 kg, an extension {tex}l{/tex} = 0.25 mm in the length of the wire is observed. Quantities {tex}d{/tex} and {tex}l{/tex} are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the {tex}Y{/tex} measurement

due to the errors in the measurements of {tex} d {/tex} and {tex} l {/tex} are the same

B

due to the error in the measurement of {tex} d {/tex} is twice that due to the error in the measurement of {tex} l {/tex}

C

due to the error in the measurement of {tex} l {/tex} is twice that due to the error in the measurement of {tex} d {/tex}

D

due to the error in the measurement of {tex} d {/tex} is four times that due to the error in the measurement of {tex} l {/tex}

##### Explanation

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Q 5. There are two Vernier calipers both of which have {tex} 1\ \mathrm { cm } {/tex} divided into {tex}10{/tex} equal divisions on the main scale. The Vernier scale of one of the calipers {tex} \left( \mathrm { C } _ { 1 } \right) {/tex} has {tex}10{/tex} equal divisions that correspond to {tex}9{/tex} main scale divisions. The Vernier scale of the other caliper {tex} \left( \mathrm { C } _ { 2 } \right) {/tex} has {tex}10{/tex} equal divisions that correspond to {tex}11{/tex} main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in {tex} \mathrm { cm } {/tex} ) by calipers {tex} \mathrm { C } _ { 1 } {/tex} and {tex} \mathrm { C } _ { 2 } {/tex}, respectively, are

A

{tex}2.85{/tex} and {tex}2.82{/tex}

{tex}2.87{/tex} and {tex}2.83{/tex}

C

{tex}2.87{/tex} and {tex}2.86{/tex}

D

{tex}2.87{/tex} and {tex}2.87{/tex}

##### Explanation

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Q 6. Using the expression {tex} 2 \mathrm { d } \sin \theta = \lambda , {/tex} one calculates the values of {tex} \mathrm { d }{/tex} by measuring the corresponding angles {tex} \theta {/tex} in the range 0 to 90{tex} ^ { \circ } . {/tex} The wavelength {tex} \lambda {/tex} is exactly known and the error in {tex} \theta {/tex} is constant for all values of {tex} \theta . {/tex} As {tex} \theta {/tex} increases from {tex} 0 ^ { \circ } {/tex}

A

The absolute error in {tex} d {/tex} remains constant

B

The absolute error in {tex} d {/tex} increases

C

The fractional error in {tex} d {/tex} remains constant

The fractional error in {tex} d {/tex} decreases