Properties of Matter — Page 9 | JEE Physics

Q81

An object is located at 2 km beneath the surface of the water. If the fractional compression

{{\Delta V} \over V}

is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ____________. [Given : density of water is 1000 kgm

-

3 and g = 9.8 ms

-

2]

A 1.44

\times

107 Nm

-

2

B 1.44

\times

109 Nm

-

2

C 1.96

\times

107 Nm

-

2

D 2.26

\times

109 Nm

-

2

Correct Answer

Option B

Solution

\beta = {{\Delta p} \over {{{\Delta V} \over V}}}

$\Rightarrow$

\beta = {{\Delta \rho gh} \over {{{\Delta V} \over V}}} = {{1000 \times 9.8 \times 2 \times {{10}^3}} \over {{{1.36} \over {100}}}}

$\Rightarrow$ $\beta$ = 1.44 $\times$ 109 N/m2

Q82

The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is :

A

{{{l_1} + {l_2}} \over 2}

B

\sqrt {{T_1}{T_2}{l_1}{l_2}}

C

{{{T_1}{l_2} - {T_2}{l_1}} \over {{T_1} - {T_2}}}

D

{{{T_1}{l_1} - {T_2}{l_2}} \over {{T_1} - {T_2}}}

Correct Answer

Option C

Solution

Suppose, I0 be the actual length of metal wire and Y be its Young's modulus. From Hooke's law,

Y = {{T{I_0}} \over {A\Delta I}}

where,

\Delta I = I - {I_0}

\Rightarrow Y = {{T{I_0}} \over {A(I - {I_0})}}

or

I - I = {{T{I_0}} \over {AY}}

$\therefore$

{{{I_1} - {I_0}} \over {{I_2} - {I_0}}} = {{{T_1}{I_0}} \over {AY}} \times {{AY} \over {{T_2}{I_0}}} = {{{T_1}} \over {{T_2}}}

\Rightarrow {I_1}{T_2} - {I_0}{T_2} = {I_2}{T_1} - {I_0}{T_1}

\Rightarrow {I_0} = {{{I_1}{T_2} - {I_2}{T_1}} \over {{T_2} - {T_1}}} = {{{T_1}{I_2} - {T_2}{I_1}} \over {({T_1} - {T_2})}}

Q83

The length of a metal wire is l1, when the tension in it is T1 and is l2 when the tension is T2. The natural length of the wire is :

A

\sqrt {{l_1}{l_2}}

B

{{{l_1}{T_2} - {l_2}{T_1}} \over {{T_2} - {T_1}}}

C

{{{l_1}{T_2} + {l_2}{T_1}} \over {{T_2} + {T_1}}}

D

{{{l_1} + {l_2}} \over 2}

Correct Answer

Option B

Solution

{T_1} = k({l_1} - {l_0})

{T_2} = k({l_2} - {l_0})

{{{T_1}} \over {{T_2}}} = {{{l_1} - {l_0}} \over {{l_2} - {l_0}}}

{{{T_1}{l_2} - {T_2}{l_1}} \over {{T_1} - {T_2}}} = {l_0}

Q84

Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is :

A

{2^{{1 \over 3}}}:1

B

1:{2^{{1 \over 3}}}

C 2 : 1

D 1 : 2

Correct Answer

Option A

Solution

The volume of a sphere is given by $\dfrac{4}{3}\pi R^3$ .

So the volume of the two small mercury drops each of radius $R$ is $2\times \dfrac{4}{3}\pi R^3$ .

When they coalesce to form a larger drop, the volume is conserved.

So, the volume of the larger drop is also $2\times \dfrac{4}{3}\pi R^3$ .

Let's denote the radius of this large drop as $R'$ .

Therefore, $2\times \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi {R'}^3$ .

Solving for $R'$ , we get $R' = 2^{1/3}R$ .

Now, the surface energy of a sphere is proportional to its surface area, and the surface area of a sphere is given by $4\pi R^2$ .

So, the ratio of total surface energy before and after the change is:

\frac{{2\times 4\pi R^2}}{{4\pi (2^{1/3}R)^2}} = \frac{{2}}{{2^{2/3}}} = {2^{1/3}}:1

Q85

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :

A 2.8

\times

10

-

4 J

B 1.5

\times

10

-

3 J

C 1.9

\times

10

-

4 J

D 9.4

\times

10

-

5 J

Correct Answer

Option A

Solution

r' = {r \over 4}

\Rightarrow \Delta E = T(\Delta S)

= T \times 4\pi (nr{'^2} - {r^2}),\,n = 64

= T \times 4\pi \times (4 - 1){r^2}

\Rightarrow \Delta E = 0.075 \times 4 \times 3.142(3) \times {10^{ - 4}}\,

J

= 2.8 \times {10^{ - 4}}

J

Q86

Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are Y1 and Y2. The combination behaves as a single wire then its Young's modulus is :

A

Y = {{2{Y_1}{Y_2}} \over {3({Y_1} + {Y_2})}}

B

Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}

C

Y = {{{Y_1}{Y_2}} \over {2({Y_1} + {Y_2})}}

D

Y = {{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}

Correct Answer

Option B

Solution

In series combination

\Delta

l = l1 + l2

Y = {{F/A} \over {\Delta l/l}} \Rightarrow \Delta l = {{Fl} \over {AY}}

\Rightarrow \Delta l \propto {l \over Y}

Equivalent length of rod after joining is = 2l As, lengths are same and force is also same in series

\Delta l = \Delta {l_1} + \Delta {l_2}

{{{l_{eq}}} \over {{Y_{eq}}}} = {l \over {{Y_1}}} + {l \over {{Y_2}}} \Rightarrow {{2l} \over Y} = {l \over {{Y_1}}} + {l \over {{Y_2}}}

$\therefore$

Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}

Q87

Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to :

A

{{{r_1}{r_2}} \over {{r_1} + {r_2}}}

B

\sqrt {{r_1}{r_2}}

C

\sqrt {r_1^2 + r_2^2}

D

{{{r_1} + {r_2}} \over 2}

Correct Answer

Option C

Solution

no. of moles is conserved n1 + n2 = n3 P1V1 + P2V2 = P3V

{{4S} \over {{r_1}}}\left( {{4 \over 3}\pi r_1^3} \right) + {{4S} \over {{r_2}}}\left( {{4 \over 3}\pi r_2^3} \right) = {{4S} \over {{r_3}}}\left( {{4 \over 3}\pi r_3^3} \right)

r_1^2 + r_2^2 = r_3^2

{r_3} = \sqrt {r_1^2 + r_2^2}

Q88

A body takes 4 min. to cool from 61

^\circ

C to 59

^\circ

C. If the temperature of the surroundings is 30

^\circ

C, the time taken by the body to cool from 51

^\circ

C to 49

^\circ

C is :

A 4 min.

B 3 min.

C 8 min.

D 6 min.

Correct Answer

Option D

Solution

{{\Delta T} \over {\Delta t}} = K({T_t} - {T_s})

Tt = average temp. T = surrounding temp.

{{61 - 59} \over 4} = K\left( {{{61 + 59} \over 2} - 30} \right)

..... (1)

{{51 - 49} \over t} = K\left( {{{51 + 49} \over 2} - 30} \right)

..... (2) Divide (1) & (2)

{t \over 4} = {{60 - 30} \over {50 - 30}} = {{30} \over {20}}

So, t = 6 minutes.

Q89

A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is : [Density of water fw = 1000 kg m

-

3 and Density of air fa = 1.2 kg m

-

3, g = 10 m/s2, Coefficient of viscosity of air = 1.8

\times

10

-

5 Nsm

-

2]

A 250.6 ms

-

1

B 43.56 ms

-

1

C 4.94 ms

-

1

D 14.4 ms

-

1

Correct Answer

Option C

Solution

At terminal speed a = 0 Fnet = 0 mg = Fv = 6 $\pi$ $\eta$ Rv

v = {{mg} \over {6\pi \eta Rv}}

v = {{{\rho _w}{{4\pi } \over 3}{R^3}g} \over {6\pi \eta R}}

= {{2{\rho _w}{R^2}g} \over {9\eta }}

= {{400} \over {81}}

m/s = 4.94 m/s

Q90

Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water T = 7.3

\times

10

-

2 Nm

-

1, angle of contact = 0, g = 10 ms2 and density of water = 1.0

\times

103 kg m

-

3]

A 3.62 mm

B 2.19 mm

C 5.34 mm

D 4.97 mm

Correct Answer

Option B

Solution

Image We have, PA = PB. [Points A & B at same horizontal level] $\therefore$

{P_{atm}} - {{2T} \over {{r_1}}} + \rho g(x + \Delta h) = {P_{atm}} - {{2T} \over {{r_2}}} + \rho gx

$\therefore$

\rho g\Delta h = 2T\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right]

= 2 \times 7.3 \times {10^{ - 2}}\left[ {{1 \over {2.5 \times {{10}^{ - 3}}}} - {1 \over {4 \times {{10}^{ - 3}}}}} \right]

$\therefore$

\Delta h = {{2 \times 7.3 \times {{10}^{ - 2}} \times {{10}^3}} \over {{{10}^3} \times 10}}\left[ {{1 \over {2.5}} - {1 \over 4}} \right]

= 2.19 $\times$ 10 $-$ 3 m = 2.19 mm Hence, option (b).