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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 $$\mu Bb)\Delta a 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Offline) A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75oC. T is given by: (Given : room temperature = 30oC, specific heat of copper = 0.1 cal/gmoC) A 825oC B 800oC C 885oC D 1250oC ## Explanation According to principle of calorimetry, Heat lost = Heat gain 100 × 0.1( – 75) = 100 × 0.1 × 45 + 170 × 1 × 45 10 – 750 = 450 + 7650 10 = 1200 + 7650 = 8850 T = 885°C 3 MCQ (Single Correct Answer) ### JEE Main 2017 (Offline) A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of: A {1 \over {81}} B 9 C {1 \over {9}} D 81 ## Explanation Stress = {{Force} \over {Area}} = {{mg} \over A} = {{\rho Vg} \over A} \left(As\ {\rho = {m \over V}} \right) \Rightarrow Stress \propto {{{L^3}} \over {{L^2}}} \Rightarrow Stress \propto L As linear dimension increases by a factor of 9, so stress also increases by a factor of 9. 4 MCQ (Single Correct Answer) ### JEE Main 2017 (Offline) An external pressure P is applied on a cube at 0oC so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and \alpha is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by: A {P \over {3\alpha K}} B {P \over {\alpha K}} C {3 \alpha \over {P K}} D 3PK\alpha ## Explanation As we know, Bulk modulus K = {{\Delta P} \over {\left( {{{ - \Delta V} \over V}} \right)}} \Rightarrow {{{\Delta V} \over V} = {P \over K}} V = V0(1 + \gamma$$\Delta$t)

${{{\Delta V} \over {{V_0}}} = \gamma \Delta t}$

$\therefore$ ${{P \over K} = \gamma \Delta t}$

$\Rightarrow$ ${\Delta t = {P \over {\gamma K}}}$ = ${{P \over {3\alpha K}}}$