Properties of Matter — Page 7 | JEE Physics

Q61

Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is :

A 4 : 1

B 8 : 1

C 2 : 1

D 0.8 : 1

Correct Answer

Option B

Solution

{P_{in}} = {P_0} + {{4T} \over {{R_1}}}

\Rightarrow 1.01 = 1 + {{4T} \over {{R_1}}}

\Rightarrow {{4T} \over {{R_1}}} = 0.01

1.02 = 1 + {{4T} \over {{R_2}}}

\Rightarrow {{4T} \over {{R_2}}} = 0.02

\therefore {{{R_2}} \over {{R_1}}} = {1 \over 2}

\Rightarrow {R_1} = 2{R_2}

{{{V_1}} \over {{V_2}}} = {{R_1^3} \over {R_2^3}} = {{8R_2^3} \over {R_2^3}} = {8 \over 1}

Q62

A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms–1. Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to:

A 104 Nm–2

B 106 Nm–2

C 108 Nm–2

D 103 Nm–2

Correct Answer

Option B

Solution

When rubber cord is stretched, then it stores potential energy and when released, this potential energy is given to the stone as kinetic energy.

So, potential energy of stretched cord = kinetic energy of stone

\Rightarrow {1 \over 2}Y{\left( {{{\Delta L} \over L}} \right)^2}A\,.\,L = {1 \over 2}m{v^2}

Here,

\Delta

L = 20 cm = 0.2 m, L = 42 cm = 0.42 m, v = 20 ms $-$ 1, m = 0.02 kg, d = 6 mm = 6 $\times$ 10 $-$ 3 m $\therefore$

A = \pi {r^2} = \pi {\left( {{d \over 2}} \right)^2} = \pi {\left( {{{6 \times {{10}^{ - 3}}} \over 2}} \right)^2}

= \pi {(3 \times {10^{ - 3}})^2} = 9\pi \times {10^{ - 6}}{m^2}

On substituting values, we get

Y = {{m{v^2}L} \over {A{{(\Delta L)}^2}}} = {{0.02 \times {{(20)}^2} \times 0.42} \over {9\pi \times {{10}^{ - 6}} \times {{(0.2)}^2}}}

\approx 3.0 \times {10^6}N{m^{ - 2}}

So, the closest value of Young's modulus is 106 Nm $-$ 2.

Q63

Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of : (density of water = 1000 kg/m3, coefficient of viscosity of water = 1mPas)

A 106

B 104

C 103

D 102

Correct Answer

Option B

Solution

Flow rate of water (Q) = 100 lit/min =

{{100 \times {{10}^{ - 3}}} \over {60}} = {5 \over 3} \times {10^{ - 3}}{m^3}

$\therefore$ Velocity of flow (v) =

{Q \over A} = {{5 \times {{10}^{ - 3}}} \over {3 \times \pi \times {{(5 \times {{10}^{ - 2}})}^2}}}

= {{10} \over {15\pi }} = {2 \over {3\pi }}\,m/s

= 0.2 m/s $\therefore$ Reynold number (Re) =

{{Dv\rho } \over \eta }

= {{\left( {10 \times {{10}^{ - 2}}} \right) \times {2 \over {3\pi }} \times 1000} \over 1} = 2 \times {10^4}

$\therefore$ Order of Re = 104

Q64

A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1 p ms–2, what will be the tensile stress that would be developed in the wire ?

A 3.1 × 106 Nm–2

B 6.2 × 106 Nm–2

C 4.8 × 106 Nm–2

D 5.2 × 106 Nm–2

Correct Answer

Option A

Solution

Tensile stress in wire will be =

{{Tensile{\rm{ }}force} \over {Cross{\rm{ }}section{\rm{ }}Area}}

=

{{mg} \over {\pi {R^2}}} = {{4 \times 3.1\pi } \over {\pi \times 4 \times {{10}^{ - 6}}}}N{m^{ - 2}}

= 3.1 × 106 Nm–2

Q65

A fluid is flowing through a horizontal pipe of varying cross-section, with speed v ms–1 at a point where the pressure is P pascal. At another point where pressure is

{P \over 2}

Pascal its speed is V ms–1. If the density of the fluid is

\rho

kg m–3 and the flow is streamline, then V is equal to :

A

\sqrt {{P \over {2\rho }} + {v^2}}

B

\sqrt {{P \over \rho } + {v^2}}

C

\sqrt {{{2P} \over \rho } + {v^2}}

D

\sqrt {{P \over \rho } + {v}}

Correct Answer

Option B

Solution

From Bernoulli's equation, P +

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

=

{P \over 2} + {1 \over 2}\rho {V^2}

$\Rightarrow$ V =

\sqrt {{P \over \rho } + {v^2}}

Q66

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other, x2 . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

A gdS(x2 + x1)2

B gdS

\left( {x_2^2 + x_1^2} \right)

C

{1 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}

D

{3 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}

Correct Answer

Option C

Solution

{u_i} = \left[ {dS{x_1}.{{{x_1}} \over 2} + dS{x_2}.{{{x_2}} \over 2}} \right]g\left\{ {dS{x_1} \to m,\,{{{x_1}} \over 2} \to h(C.O.M)} \right\}

Here total volume remains same. $\therefore$ Vi = Vf $\Rightarrow$ S(x1 + x2) = S(h + h) $\Rightarrow$ h =

{{{x_1} + {x_2}} \over 2}

uf = (dSh)g

{h \over 2} \times 2

$\Rightarrow$

{u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} \right) \times \left( {{{{x_1} + {x_2}} \over 4}} \right) \times 2} \right]g

$\therefore$

{u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} \right)}^2}} \over 4}} \right]

= dsg{{{{\left( {{x_1} - {x_2}} \right)}^2}} \over 4}

Q67

A cube of metal is subjected to a hydrostatic pressure of 4 GPa. The percentage change in the length of the side of the cube is close to : (Given bulk modulus of metal, B = 8

\times

1010 Pa)

A 0.6

B 20

C 1.67

D 5

Correct Answer

Option C

Solution

Bulk Modulus, B =

\left( - \right){{\Delta P} \over {\Delta V/V}}

\Delta P = -\left( {{{\Delta V} \over V}} \right).B

= -{{3\Delta L} \over L} \times B

$\therefore$

|{{\Delta L} \over L}| = {{\Delta P} \over {3B}}

$\therefore$ % change,

{{\Delta L} \over L} \times 100\%

=

{1 \over 3}{{\Delta P} \over B} \times 100

=

{{4 \times {{10}^9}} \over {8 \times {{10}^{10}}}} \times 100

=

{1 \over {60}} \times 100

= 1.67

Q68

A metallic sphere cools from 50oC to 40o in 300 s. If atmospheric temperature around is 20oC, then the sphere’s temperature after the next 5 minutes will be close to :

A 35oC

B 31oC

C 33oC

D 28oC

Correct Answer

Option C

Solution

{{\Delta T} \over {\Delta t}} = k\left( {{{{T_f} + {T_i}} \over 2} - {T_0}} \right)

$\Rightarrow$

{{50 - 40} \over {300}} = k\left( {{{90} \over 2} - 20} \right)

$\Rightarrow$

{{40 - T} \over {300}} = k\left( {{{40 + T} \over 2} - 20} \right)

$\Rightarrow$

{{10} \over {40 - T}} = {{25 \times 2} \over {40 + T - 40}}

$\Rightarrow$

{1 \over {40 - T}} = {5 \over T}

$\Rightarrow$

T = 200 - 5T

$\Rightarrow$

6T = 200

$\Rightarrow$

T = 33^\circ C

Q69

A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodide (surface tension = 0.05 Nm–1, density = 667 kg m–3) which rises to height h in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of 60o with one another. Then h is close to (g = 10 ms–2)

A 0.049 m

B 0.087 m

C 0.137 m

D 0.172 m

Correct Answer

Option B

Solution

h =

{{2T\cos \theta } \over {\rho gr}}

=

{{2 \times 0.05 \times {\sqrt 3 \over 2}} \over {667 \times 10 \times 0.15 \times {{10}^{ - 3}}}}

= 0.087 m

Q70

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is :

A

{3 \over 4}

B

{9 \over {16}}

C

{{\sqrt 3 } \over 2}

D

{{81} \over {256}}

Correct Answer

Option B

Solution

Using equation of continuity A1V1 = A2V2 $\Rightarrow$

{{{V_1}} \over {{V_2}}} = {{{A_2}} \over {{A_1}}}

=

{\left( {{{4.8} \over {6.4}}} \right)^2}

=

{9 \over {16}}