1

### JEE Main 2017 (Online) 8th April Morning Slot

A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $\Delta$T. The net change in its length is zero. Let $\ell$ be the length of the rod, A its area of cross-section,Y its Young’s modulus, and $\alpha$ its coefficient of linear expansion. Then, F is equal to :
A
$\ell$2 Y$\alpha$ $\Delta$T
B
$\ell$A Y$\alpha$ $\Delta$T
C
A Y$\alpha$ $\Delta$T
D
${{AY} \over {\alpha \,\Delta T}}$

## Explanation

Because of thermal expansion, change in length

($\Delta $$\ell ) = \ell \alpha \Delta T . . . . .(1) Because of compressive force, the compansion is \Delta$$\ell$ ' ,

$\therefore\,\,\,$ Young's Modulus (y) = ${{{F \over A}} \over {{{\Delta \ell} \over \ell }}}$

$\Rightarrow $$\,\,\, F = YA {{\Delta \ell '} \over \ell } As net change in length is 0 . So, \Delta \ell ' = \Delta \ell = \ell \alpha \,\Delta T \therefore\,\,\, F = YA \times {{\ell \alpha \,\Delta T} \over \ell } = AY\alpha$$\Delta$T
2

### JEE Main 2017 (Online) 9th April Morning Slot

A steel rail of length 5 m and area of cross section 40cm2 is prevented from expanding along its length while the temperature rises by 10oC. If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately :
A
2 $\times$ 107 N
B
1 $\times$ 105 N
C
2 $\times$ 109 N
D
3 $\times$ 10$-$5 N

## Explanation

Young's modulus (Y) = ${{{F \over A}} \over {{{\Delta L} \over L}}}$

as  ${{{\Delta L} \over L}}$ = $\alpha$ $\Delta $$\theta \therefore\,\,\, Y = {{F \over {A\alpha \Delta \theta }}} \Rightarrow$$\,\,\,$ F = YA$\alpha $$\Delta$$\theta$

= 2 $\times$ 1011 $\times$ 40$\times$10$-$4 $\times$ 1.2 $\times$ 10$-$5 $\times$ 10

= 9.6 $\times$ 104 N

$\simeq$ 1 $\times$ 105 N
3

### JEE Main 2017 (Online) 9th April Morning Slot

Two tubes of radii r1 and r2, and lengths l1 and l2 , respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1 and P2 are pressure differences across the two tubes.

If P2 is 4P1 and l2 is ${{{1_1}} \over 4}$, then the radius r2 will be equal to :
A
r1
B
2r1
C
4r1
D
${{{r_1}} \over 2}$

## Explanation

We know,

Rate of flow of liquid through narrow tube,

${{dv} \over {dt}}$ = ${{\pi {{\Pr }^4}} \over {8\eta l}}$

Both tubes are connected in series so rate of flow of liquid is same.

$\therefore\,\,\,$ ${{\pi {P_1}r_1^4} \over {8\eta {l_1}}}$ = ${{\pi {P_2}r_2^4} \over {8\eta {l_2}}}$

$\Rightarrow $$\,\,\, {{{P_1}{r_1}^4} \over {{l_1}}} = {{{P_2}{r_2}^4} \over {{l_2}}} Given, P2 = 4P1 and {l_2} = {{{l_1}} \over 4} \Rightarrow$$\,\,\,$ ${{{P_1}{r_1}^4} \over {{l_1}}}$ = ${{4{P_1}{r_2}^4} \over {{{{l_1}} \over 4}}}$

$\Rightarrow $$\,\,\, {r_2}^4 = {{{r_1}^4} \over {16}} \Rightarrow$$\,\,\,$ r2 = ${{{r_1}} \over 2}$
4

### JEE Main 2018 (Offline)

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left( {{dr \over r}} \right)$ is:
A
${{mg} \over {Ka}}$
B
${{Ka} \over {mg}}$
C
${{Ka} \over {3mg}}$
D
${{mg} \over {3Ka}}$

## Explanation

Because of m mass the extra pressure created is,

$\Delta$P = ${{mg} \over a}$

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

Given $\beta$ = K

$\therefore\,\,\,$ K = ${{{{mg} \over a}} \over {{{\Delta V} \over V}}}$

We know volume of sphere,

V = ${4 \over 3}\pi {r^3}$

$\therefore\,\,\,$ ${{dV} \over V}$ = 3 ${{dr} \over r}$

$\therefore\,\,\,$ K = ${{{{mg} \over a}} \over {3{{dr} \over r}}}$

$\Rightarrow$$\,\,\,$ ${{dr} \over r}$ = ${{mg} \over {3Ka}}$