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### JEE Main 2016 (Online) 10th April Morning Slot

A thin 1 m long rod has a radius of 5 mm. A force of 50 $\pi$kN is applied at one end to determine its Young’s modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?
A
${{\Delta \gamma } \over \gamma }$ gets minimum contribution from the uncertainty in the length.
B
The figure of merit is the largest for the length of the rod.
C
The maximum value of $\gamma$ that can be determined is 2 $\times$ 1014 N/m2
D
${{\Delta \gamma } \over \gamma }$ gets its maximum contribution from the uncertainty in strain
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### JEE Main 2016 (Online) 10th April Morning Slot

A bottle has an opening of radius a and length b. A cork of length b and radius (a + $\Delta$a) where ($\Delta$a < < a) is compressed to fit into the opening completely (See figure). If the bulk modulus of cork is B and frictional coefficient between the bottle and cork is $\mu$ then the force needed to push the cork into the bottle is : A
($\pi$ $\mu$ B b) $\Delta$a
B
(2$\pi$ $\mu$ B b) $\Delta$a
C
($\pi$ $\mu$ B b) a
D
(4$\pi$ $\mu$ B b) $\Delta$a

## Explanation

Bulk modulus, B  =  ${{\Delta P} \over {{{\Delta V} \over V}}}$

Vi  =  $\pi$ (a + $\Delta$a)2b

Vf  =  $\pi$a2b

$\therefore$   $\Delta$V = 2$\pi$ab$\Delta$a

$\therefore$   ${{{\Delta V} \over V}}$ = ${{2\pi ab\Delta a} \over {\pi {a^2}b}}$ = ${{2\Delta a} \over a}$

$\therefore$   $\Delta$P = B $\times$ ${{{\Delta V} \over V}}$

= B $\times$ ${{2\Delta a} \over a}$

Normal Force (N) = $\Delta$P $\times$ A

= B $\times$ ${{2\Delta a} \over a}$ $\times$ (2$\pi$a)b

=   4$\pi$B$\Delta$ab

$\therefore$   Frictional force = $\mu$N = (4$\pi$$\mu$Bb)$\Delta$a
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### JEE Main 2017 (Offline)

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75oC. T is given by: (Given : room temperature = 30oC, specific heat of copper = 0.1 cal/gmoC)
A
825oC
B
800oC
C
885oC
D
1250oC

## Explanation

According to principle of calorimetry,

Heat lost = Heat gain

100 × 0.1( – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

10 – 750 = 450 + 7650

10 = 1200 + 7650 = 8850

T = 885°C
4

### JEE Main 2017 (Offline)

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of:
A
${1 \over {81}}$
B
9
C
${1 \over {9}}$
D
81

## Explanation

Stress = ${{Force} \over {Area}}$ = ${{mg} \over A}$ = ${{\rho Vg} \over A}$ $\left(As\ {\rho = {m \over V}} \right)$

$\Rightarrow$ Stress $\propto$ ${{{L^3}} \over {{L^2}}}$

$\Rightarrow$ Stress $\propto$ L

As linear dimension increases by a factor of 9, so stress also increases by a factor of 9.