Properties of Matter — Page 8 | JEE Physics

Q71

Consider a solid sphere of radius R and mass density

\rho \left( r \right) = {\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)

,

0 < r \le R

The minimum density of a liquid in which it will float is :

A

{{2{\rho _0}} \over 3}

B

{{2{\rho _0}} \over 5}

C

{{{\rho _0}} \over 5}

D

{{{\rho _0}} \over 3}

Correct Answer

Option B

Solution

Mass of solid =

\int\limits_0^R {\rho dV}

=

\int\limits_0^R {{\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)4\pi {r^2}dr}

=

{\rho _0}4\pi \left[ {\int\limits_0^R {{r^2}dr} - {1 \over {{R^2}}}\int\limits_0^R {{r^4}dr} } \right]

=

{\rho _0}4\pi \left[ {{{{R^3}} \over 3} - {{{R^5}} \over 5}} \right]

=

{\rho _0}{{8\pi {R^3}} \over {15}}

For minimum density of liquid, solid sphere has to float (completely immersed) in the liquid.

So Weight of the sphere = Buoyant force mg = Fb $\Rightarrow$

{\rho _0}{{8\pi {R^3}} \over {15}}

g =

{\rho _l}{4 \over 3}\pi {R^3}g

$\Rightarrow$

{\rho _l} = {{2{\rho _0}} \over 5}

Q72

A leak proof cylinder of length 1m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C, relative to the density at T = 0°C is close to :

A 1.04

B 1.26

C 1.01

D 1.03

Correct Answer

Option C

Solution

In both cases weight of the cylinder is balanced by buoyant force exerted by liquid. Case - 1 :

{\rho _{0^\circ C}}\left( {100 - 20} \right) \times g

= mg Case - 2 :

{\rho _{4^\circ C}}\left( {100 - 21} \right) \times g

= mg $\therefore$

{{{\rho _{4^\circ C}}\left( {100 - 21} \right)} \over {{\rho _{0^\circ C}}\left( {100 - 20} \right)}}

=1 $\Rightarrow$

{{{\rho _{4^\circ C}}} \over {{\rho _{0^\circ C}}}}

=

{{80} \over {79}}

= 1.01

Q73

A small spherical droplet of density d is floating exactly half immersed in a liquid of density

\rho

and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet) :

A

r = \sqrt {{T \over {\left( {d - \rho } \right)g}}}

B

r = \sqrt {{{2T} \over {3\left( {d + \rho } \right)g}}}

C

r = \sqrt {{T \over {\left( {d + \rho } \right)g}}}

D

r = \sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}}

Correct Answer

Option D

Solution

T.2\pi r + {2 \over 3}\pi {r^3}\rho g = {4 \over 3}\pi {r^3}dg

$\Rightarrow$ T =

{{{r^2}} \over 3}\left( {2d - \rho } \right)g

$\Rightarrow$ r =

\sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}}

Q74

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is:

A 1 : 2

B 2 : 1

C

1:\sqrt 2

D

\sqrt 2 :1

Correct Answer

Option D

Solution

{{du} \over {dv}}

=

{1 \over 2}

$\times$ stress × strain =

{1 \over 2}{F \over A} \times {F \over {AY}}

$\propto$

{1 \over {{A^2}}}

$\propto$

{1 \over {{d^4}}}

{{du} \over {dv}}

=

{1 \over 4}

$\Rightarrow$

{\left( {{{{d_1}} \over {{d_2}}}} \right)^4}

= 4 $\Rightarrow$

{{{d_1}} \over {{d_2}}} = {\left( 4 \right)^{{1 \over 4}}}

=

\sqrt 2 :1

Q75

If Y, K and

\eta

are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.

A

Y = {{9K\eta } \over {3K - \eta }}N/{m^2}

B

Y = {{9K\eta } \over {2\eta + 3K}}N/{m^2}

C

\eta = {{3YK} \over {9K + Y}}N/{m^2}

D

K = {{Y\eta } \over {9\eta - 3Y}}N/{m^2}

Correct Answer

Option D

Solution

We know that,

Y = 3K(1 - 2\sigma )

\Rightarrow \sigma = {1 \over 2}\left( {1 - {Y \over {3K}}} \right)

..... (i) Also,

Y = 2\eta (1 + \sigma )

\Rightarrow \sigma = {Y \over {2\eta }} - 1

.... (ii) On comparing Eqs. (i) and (ii), we get

\left( {1 - {Y \over {3K}}} \right){1 \over 2} = {Y \over {2\eta }} - 1

On solving, we get

K = {{\eta Y} \over {9\eta - 3Y}}

N/m2

Q76

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be :

A

{{2T} \over J}\left( {{1 \over r} - {1 \over R}} \right)

B

{{3T} \over J}\left( {{1 \over r} - {1 \over R}} \right)

C

{{3T} \over rJ}

D

{{2T} \over rJ}

Correct Answer

Option B

Solution

R is the radius of bigger drop. r is the radius of n water drops.

Water drops are combined to make bigger drop.

So, Volume of n drops = volume of bigger drop

n\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}

$\Rightarrow$

R = r{n^{1/3}} \Rightarrow n = {\left( {{R \over r}} \right)^3}

Loss in surface energy,

\Delta

U = T $\times$ (Change in surface area)

\Delta

U = T (n4 $\pi$ r2 $-$ 4 $\pi$ R2)

\Delta U = 4\pi T\left[ {{{\left( {{R \over r}} \right)}^3}{r^2} - {R^2}} \right] = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over J}

$\therefore$

{{\Delta U} \over V} = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over {J \times {4 \over 3}\pi {R^3}}} = {{3T} \over J}\left[ {{1 \over r} - {1 \over R}} \right]

Q77

The length of metallic wire is l1 when tension in it is T1. It is l2 when the tension is T2. The original length of the wire will be :

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

Assuming Hooke's law to be valid.

T \propto (\Delta l)

T = k(\Delta l)

Let, l0 = natural length (original length)

\Rightarrow T = k(l - {l_0})

so,

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

&

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

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

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

Q78

The pressure acting on a submarine is 3

\times

105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is 1

\times

105 Pa density of water is 103 kg m

-

3, g = 10 ms

-

2)

A

{{200} \over 5}

%

B

{{200} \over 3}

%

C

{{3} \over 200}

%

D

{{5} \over 200}

%

Correct Answer

Option B

Solution

P = P0 + h $\rho$ g = 3 $\times$ 105 Pa $\Rightarrow$ h $\rho$ g = 3 $\times$ 105 $-$ 1 $\times$ 105 $\Rightarrow$ h $\rho$ g = 2 $\times$ 105 $\therefore$ 2h $\rho$ g = 4 $\times$ 105 $\therefore$ P' = P0 + 4 $\times$ 105 $\therefore$ P' = 5 $\times$ 105 Pa $\therefore$ % increase in pressure =

{{P' - P} \over P} \times 100

= {{(5 - 3) \times {{10}^5}} \over {3 \times {{10}^5}}} \times 100

= {{200} \over 3}

%

Q79

What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10

-

3 Pa s, respectively. (Density of water : 103 kg/m3)

A Steady flow to unsteady flow

B Unsteady to steady flow

C Remains turbulent flow

D Remains steady flow

Correct Answer

Option A

Solution

The nature of flow is determined by reynolds no

R = {{\rho VD} \over \eta }

If R < 1000 $\to$ flow is steady 1000 < R < 2000 $\to$ flow becomes unsteady R > 2000 $\to$ flow is turbulent

{R_1} = {{4 \times {{10}^3} \times 0.18 \times {{10}^{ - 3}}} \over {60 \times \pi \times {{10}^{ - 2}} \times {{10}^{ - 3}}}} = {{4 \times {{10}^5} \times 0.18} \over {60\pi }}

= 0.0038 \times {10^5} = 380

{R_2} = {{0.48} \over {0.18}} \times 380 = 1018

Q80

When two soap bubbles of radii a and b (b > a) coalesce, the radius of curvature of common surface is :

A

{{b - a} \over {ab}}

B

{{a + b} \over {ab}}

C

{{ab} \over {a + b}}

D

{{ab} \over {b - a}}

Correct Answer

Option D

Solution

{P_1} - {P_0} = {{4S} \over b}

....(1)

{P_2} - {P_0} - {{4S} \over a}

.....(2)

{P_2} - {P_1} = {{4S} \over R}

......(3) eq(2) - eq(1) = eq(3) $\Rightarrow$

{1 \over a} - {1 \over b} = {1 \over R}

$\therefore$

R = {{ab} \over {b - a}}