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### JEE Main 2016 (Online) 9th April Morning Slot

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is $\alpha$ then the value of S is :
A
$\alpha$
B
${\alpha \over 2}$
C
2$\alpha$
D
${1 \over \alpha }$

## Explanation

Change of length of wire with temperature,

$\Delta $$\ell = \alpha \ell \Delta \theta Time period of pendulum at temperature \theta , T\theta = 2\pi$$\sqrt {{{\ell + \Delta \ell } \over g}}$

= 2$\pi$$\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}}$

= $2\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}$

= $2\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}$

$\simeq$  T0 $\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)$

Here T0 = time period at temperature 0oC.

$\therefore$   Change in time period,

$\Delta$T = T$\theta$ $-$ T0

= ${{{T_0}\Delta \ell } \over {2\ell }}$

= ${{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}$

$\therefore$   ${{\Delta T} \over {\Delta \theta }}$ = ${{{T_0}\alpha } \over 2}$

Given that T0 = 2,

$\therefore$   ${{\Delta T} \over {\Delta \theta }}$ = ${{2\alpha } \over 2}$ = $\alpha$

${{\Delta T} \over {\Delta \theta }}$ is the shape of $\Delta$T and ${\Delta \theta }$

curve = S (given)

$\therefore$   S = $\alpha$
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### JEE Main 2016 (Online) 9th April Morning Slot

Which of the following option correctly describes the variation of the speed v and acceleration ‘a’ of a point mass falling vertically in a viscous medium that applies a force F = − kv, where ‘k’ is a constant, on the body ? (Graphs are schematic and not drawn to scale)
A B C D ## Explanation

Equation of motion for the mass,

ma = mg $-$ kv

$\Rightarrow$    ${{dv} \over {dt}} = {{mg - kv} \over m}$

$\Rightarrow$   $\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt}$

$\Rightarrow$   $- {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$

$\Rightarrow$   $\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$

$\Rightarrow$   $1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$

$\Rightarrow$   ${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$

$\Rightarrow$   $v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$

ma $=$ mg $-$ k $\times$ ${{mg} \over k}$ (1 $-$ e$^{ - {{kt} \over m}}$)

$=$  mg $-$ mg + mge$^{ - {{kt} \over m}}$

a $=$ g e$^{ - {{kt} \over m}}$
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### JEE Main 2016 (Online) 9th April Morning Slot Consider a water jar of radius R that has water filled up to height H and is kept on astand of height h (see figure). Through a hole of radius r (r << R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :
A
$x = r\left( {{H \over {H + h}}} \right)$
B
$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 2}}}$
C
$x = r{\left( {{H \over {H + h}}} \right)^{{1 \over 4}}}$
D
$x = r{\left( {{H \over {H + h}}} \right)^{{2}}}$

## Explanation

v1 = velocity of water when it leak from hole

v2 = velocity of water when it reach the ground.

From Bernoulli's principle,

${1 \over 2}\rho {v_1}^2 + \rho gh$ = ${1 \over 2}\rho {v_2}^2$

$\Rightarrow$   ${v_1}^2$ + 2gh = ${v_2}^2$

From Torricelli's theorem,

v1 = $\sqrt {2gH}$

$\therefore$   ${v_2}^2$ = 2gh + 22gH

From continuity equation,

${A_1}{v_1}$ = ${A_2}{v_2}$

$\Rightarrow$   $\pi {r^2} \times \sqrt {2gH}$ = $\pi {x^2}\sqrt {2g\left( {h + H} \right)}$

$\Rightarrow$   x2  =  r2$\sqrt {{H \over {H + g}}}$

$\Rightarrow$   x = r ${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$
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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