Properties of Matter — Page 6 | JEE Physics

Q51

Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1 . The cross-sectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be : (Take g = 10 ms–2)

A 5 × 10–4 m2

B 2 × 10–5 m2

C 5 × 10–5 m2

D 1 × 10–5 m2

Correct Answer

Option C

Solution

Using Bernoullie’s equation

{v_2} = \sqrt {v_1^2 + 2gh}

Equation of continuity A1V1 = A2V2 (1 cm3)(1m/s) =

\left( {{A_2}} \right)\left( {\sqrt {{{\left( 1 \right)}^2} + 2 \times 10 \times {{15} \over {100}}} } \right)

\Rightarrow {A_2}\left( {\ln c{m^2}} \right) = {1 \over 2}

\Rightarrow {A_2} = 5 \times {10^{ - 5}}{m^2}

Q52

A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 1/16 th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is :

A

2T\sqrt {{1 \over {10}}}

B

4T\sqrt {{1 \over {14}}}

C

4T\sqrt {{1 \over {15}}}

D

2T\sqrt {{1 \over {14}}}

Correct Answer

Option C

Solution

For a simple pendulum T =

2\pi \sqrt {{L \over {{g_{err}}}}}

Situation 1: when pendulum is in air $\to$ geff = g Situation 2:when pendulum is in liquid $\to$ geff =

\left( {1 - {{{\rho _{liquid}}} \over {{\rho _{body}}}}} \right) = g\left( {1 - {1 \over {16}}} \right) = {{15g} \over {16}}

So,

{{{T^{'}}} \over T} = {{2\pi \sqrt {{L \over {15g/16}}} } \over {2\pi \sqrt {{L \over g}} }}

\Rightarrow {T^{'}} = {{4T} \over {\sqrt {15} }}

Q53

A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 106 Pa. Then d2 –d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2 ) :

A 600 m

B 400 m

C 300 m

D 500 m

Correct Answer

Option C

Solution

The pressure experienced by a submarine at a certain depth in the sea is given by the formula: $P = \rho g h$ where: $P$ is the pressure $\rho$ is the density of the fluid (sea water in this case) $g$ is the acceleration due to gravity $h$ is the height (or depth in this case) Given: $\rho = 10^3 \, kg/m^3$ $g = 10 \, m/s^2$ We are looking for the difference in depth, $d_2 - d_1$ , which corresponds to the difference in pressure $\Delta P$ : $\Delta P = P_2 - P_1 = \rho g (d_2 - d_1)$ Rearranging the above equation, we get: $d_2 - d_1 = \dfrac{\Delta P}{\rho g}$ Given: $P_2 = 8.08 \times 10^6 \, Pa$ $P_1 = 5.05 \times 10^6 \, Pa$ $\Delta P = P_2 - P_1 = 8.08 \times 10^6 \, Pa - 5.05 \times 10^6 \, Pa = 3.03 \times 10^6 \, Pa$ So, $d_2 - d_1 = \dfrac{3.03 \times 10^6 \, Pa}{10^3 \, kg/m^3 \times 10 \, m/s^2} = 303 \, m$ The closest answer among the options provided is Option C, 300 m.

Q54

In an experiment, brass and steel wires of length 1 m each with areas of cross section 1mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young's Modulus for steel and brass are, respectively, 120 × 109 N/m2 and 60 × 109 N/m2]

A 8.0 × 106 N/m2

B 1.2 × 106 N/m2

C 0.2 × 106 N/m2

D 1.8 × 106 N/m2

Correct Answer

Option A

Solution

Corresponding to the stress ( $\sigma$ ) Total elongation

\Delta {I_{net}} = {{\sigma {L_1}} \over {{Y_1}}} + {{\sigma {L_2}} \over {{Y_2}}}

\sigma = \Delta I\left( {{{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}} \right)

= 0.2 \times {10^{ - 3}} \times \left( {{{120 \times 60} \over {180}}} \right) \times {10^9}

= 8 \times {10^6}{N \over {{m^2}}}

Q55

A solid sphere, of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity . The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, v2, when falling through the same fluid, the ratio (v1/v2) equals :

A

{1 \over 9}

B

{1 \over {27}}

C 27

D 9

Correct Answer

Option D

Solution

{4 \over 3}

$\pi$ R3 = 27 $\times$

{4 \over 3}

$\pi$ r3 $\Rightarrow$ r =

{R \over 3}

Terminal velocity,

{V_T} = {2 \over 9}{{{r^2}} \over \eta }\left( {{\sigma _s} - {\rho _l}} \right)g

$\therefore$ VT $\propto$ r2 $\therefore$

{{{v_1}} \over {{v_2}}} = {{{R^2}} \over {{r^2}}}

=

{{{R^2}} \over {{{{R^2}} \over 9}}}

= 9

Q56

The number density of molecules of a gas depends on their distance r from the origin as,

n\left( r \right) = {n_0}{e^{ - \alpha {r^4}}}

. Then the total number of molecules is proportional to :

A

{n_0}{\alpha ^{ - 3/4}}

B

{n_0}{\alpha ^{ - 3}}

C

{n_0}{\alpha ^{1/4}}

D

\sqrt {{n_0}} {\alpha ^{1/2}}

Correct Answer

Option A

Solution

Lets take an element hollow sphere of thickness dr Vol. of element dV = 4 $\pi$ r2dr Total number of molecules, N =

\int\limits_0^\infty {n\,dV}

=

\int\limits_0^\infty {{n_0}{e^{ - \alpha {r^4}}}\,4\pi {r^2}dr}

Let

{{e^{ - \alpha {r^4}}}}

= t ......................(1) $\therefore$

- 4\alpha {e^{ - \alpha {r^4}}}{r^3}dr

= dt $\Rightarrow$

{{dt} \over { - 4\alpha r}} = {e^{ - \alpha {r^4}}}{r^2}dr

.....................(2) Taking

\ln

to the both sides of the equation (1), we get

{\ln}\left( t \right) = - \alpha {r^4}

$\Rightarrow$

r = {\left( {{{\ln t} \over { - \alpha }}} \right)^{{1 \over 4}}}

.......(3) Putting this value of r in equation (2),

{{dt} \over { - 4\alpha {{\left( {{{\ln t} \over { - \alpha }}} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr

$\Rightarrow$

{{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}} = {e^{ - \alpha {r^4}}}{r^2}dr

Putting in the integration, we get N =

{n_0}4\pi \int\limits_1^0 {{{{\alpha ^{{{ - 3} \over 4}}}dt} \over { - 4{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}}

=

- {n_0}{\alpha ^{{{ - 3} \over 4}}}\pi \int\limits_1^0 {{{dt} \over {{{\left( { - \ln t} \right)}^{{1 \over 4}}}}}}

$\therefore$ N $\propto$

{n_0}{\alpha ^{{{ - 3} \over 4}}}

Q57

A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to :

A

{{3F} \over {\left( {\pi {r^2}YT} \right)}}

B

{{6F} \over {\left( {\pi {r^2}YT} \right)}}

C

{F \over {\left( {3\pi {r^2}YT} \right)}}

D

{9F\left( {\pi {r^2}YT} \right)}

Correct Answer

Option A

Solution

Change in length due to temperature change,

\Delta

l

=

l

$\alpha$

\Delta

T

{{\Delta l} \over l}

= $\alpha$ T [ Here

\Delta

T = T ] Y =

{{{F \over {\pi {r^2}}}} \over {{{\Delta l} \over l}}}

=

{{{F \over {\pi {r^2}}}} \over {\alpha T}}

$\Rightarrow$ Y =

{F \over {\pi {r^2}\alpha T}}

$\Rightarrow$ $\alpha$ =

{F \over {\pi {r^2}YT}}

We know, The coefficient of volume expansion ( $\gamma$ ) = 3 $\alpha$ $\therefore$ $\gamma$ =

{{3F} \over {\pi {r^2}YT}}

Q58

An air bubble of negligible weight having radius r rises steadily through a solution of density

\sigma

at speed v. The coefficient of viscosity of the solution is given by :

A

\eta = {{4r\sigma g} \over {9v}}

B

\eta = {{2{r^2}\sigma g} \over {9v}}

C

\eta = {{2\pi {r^2}\sigma g} \over {9v}}

D

\eta = {{2{r^2}\sigma g} \over {3\pi v}}

Correct Answer

Option B

Solution

Air bubble moves with constant speed v. So net force = 0. $\therefore$ Buoyant Force = Viscous force

\Rightarrow {F_b} = {F_v}

\Rightarrow \sigma \times {4 \over 3}\pi {r^3}g = 6\pi nrv

\Rightarrow n = {{2\sigma {r^2}g} \over {9v}}

Q59

The top of a water tank is open to air and its water level is mainted. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the center of the opening from the level of water in the tank is close to :

A 6.0 m

B 4.8 m

C 9.6 m

D 2.9 m

Correct Answer

Option B

Solution

Here water level is kept same all the time.

So, the amount of water remove from the hole put in the tank the top to keep the water level same.

$\therefore$ Inflow volume role = outflow volume $\Rightarrow$

{{0.74} \over {60}} = Av

$\Rightarrow$

{{0.74} \over {60}} = \pi {r^2}\left( {\sqrt {2gh} } \right)

$\Rightarrow$

{{0.74} \over {60}} = \pi \left( {4 \times {{10}^{ - 4}}} \right) \times \sqrt {2gh}

$\Rightarrow$

\sqrt {2gh}

=

{{740} \over {24\pi }}

$\Rightarrow$ 2gh =

{{740 \times 740} \over {24 \times 24 \times {\pi ^2}}}

$\Rightarrow$ h = 4.8 m

Q60

A liquid of density

\rho

is coming out of a hose pipe of radius a with horizontal speed

\upsilon

and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be :

A

{3 \over 4}\rho {v^2}

B

{1 \over 4}\rho {v^2}

C

{1 \over 2}\rho {v^2}

D

\rho {v^2}

Correct Answer

Option A

Solution

Momentum per second carried by liquid per second is $\rho$ av2 net force due to reflected liquid = 2 $\times$

\left[ {{1 \over 4}\rho a{v^2}} \right]

net force due to stopped liquid =

{{1 \over 4}\rho a{v^2}}

Total force =

{{3 \over 4}\rho a{v^2}}

net pressure =

{{3 \over 4}\rho {v^2}}