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Properties of Matter

JEE Physics · 167 questions · Page 3 of 17 · Click an option or "Show Solution" to reveal answer

Q21
Work done in increasing the size of a soap bubble from a radius of 33 cmcm to 55 cmcm is nearly (Surface tension of soap solution =0.03Nm−1, = 0.03N{m^{ - 1}},
A 0.2πmJ0.2\pi mJ
B 2πmJ2\pi mJ
C 0.4πmJ0.4\pi mJ
D 4πmJ4\pi mJ
Correct Answer
Option C
Solution
W=T×  W = T \times \,\,

change in surface area

W=2T4π[(5)2−(3)2]×10−4W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}
=2×0.03×4π[25−9]×10−4 J= 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J
=0.4π×10−3 J= 0.4\pi \times {10^{ - 3}}\,J
=0.4πmJ= 0.4\pi mJ
Q22
A uniform heavy rod of weight 10 kg ms−-2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 ×\times 1011 Nm−-2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.
A 2 ×\times 10−-9 m
B 5 ×\times 10−-8 m
C 4 ×\times 10−-8 m
D 5 ×\times 10−-10 m
Correct Answer
Option D
Solution

We know,

Δl=WL2AY\Delta l = {{WL} \over {2AY}}
Δl=10×12×5×100×10−4×2×1011\Delta l = {{10 \times 1} \over {2 \times 5}} \times 100 \times {10^{ - 4}} \times 2 \times {10^{11}}
Δl=12×10−9=5×10−10\Delta l = {1 \over 2} \times {10^{ - 9}} = 5 \times {10^{ - 10}}

m Option (d)

Q23
An object of mass m is suspended at the end of a massless wire of length L and area of crosssection A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is :
A f=12πYAmLf = {1 \over {2\pi }}\sqrt {{{YA} \over {mL}}}
B f=12πmLYAf = {1 \over {2\pi }}\sqrt {{{mL} \over {YA}}}
C f=12πYLmAf = {1 \over {2\pi }}\sqrt {{{YL} \over {mA}}}
D f=12πmAYLf = {1 \over {2\pi }}\sqrt {{{mA} \over {YL}}}
Correct Answer
Option A
Solution

An elastic wire can be treated as a spring with k =

YAl{{YA} \over l}

T =

2πmk2\pi \sqrt {{m \over k}}

⇒\Rightarrow f =

12πkm{1 \over {2\pi }}\sqrt {{k \over m}}

=

12πYAml{1 \over {2\pi }}\sqrt {{{YA} \over {ml}}}
Q24
In an experiment to verify Stokes law, a small spherical ball of radius r and density ρ\rho falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to : (ignore viscosity of air)
A r
B r4
C r3
D r2
Correct Answer
Option B
Solution

After falling through h, the velocity be equal to terminal velocity

2gh\sqrt {2gh}

=

29r2gη(ρl−ρ){2 \over 9}{{{r^2}g} \over \eta }\left( {{\rho _l} - \rho } \right)

⇒\Rightarrow h =

281r4g(ρl−ρ)2η2{2 \over {81}}{{{r^4}g{{\left( {{\rho _l} - \rho } \right)}^2}} \over {{\eta ^2}}}

⇒\Rightarrow h ∝\propto r4

Q25
A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is 278{{27} \over 8} w.r.t water, the value of r is :
A 23{{2} \over 3}R
B 49{{4} \over 9}R
C 13{{1} \over 3}R
D 89{{8} \over 9}R
Correct Answer
Option D
Solution
43π(R3−r3)pm g=43πR3pw g{4 \over 3}\pi \left( {{R^3} - {r^3}} \right){p_m}\,g = {4 \over 3}\pi {R^3}{p_w}\,g
1−(rR)3=8271 - {\left( {{r \over R}} \right)^3} = {8 \over {27}}
⇒rR=(1927)1/3=191/33\Rightarrow {r \over R} = {\left( {{{19} \over {27}}} \right)^{1/3}} = {{{{19}^{1/3}}} \over 3}
=0.88≃89= 0.88 \simeq {8 \over 9}
Q26
A wire elongates by ll mmmm when a LOAD WW is hanged from it. If the wire goes over a pulley and two weights WW each are hung at the two ends, the elongation of the wire will be (in mmmm)
A ll
B 2l2l
C zero
D l/2l/2
Correct Answer
Option A
Solution

Case

(i)(i)

At equilibrium,

T=WT=W
Y=W/Aℓ/L              ...(1)Y = {{W/A} \over {\ell /L}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ...\left( 1 \right)

Case

(ii)(ii)

At equilibrium

T=WT=W

∴\therefore

Y=W/Aℓ/2L/2⇒Y=W/Aℓ/LY = {{W/A} \over {{{\ell /2} \over {L/2}}}} \Rightarrow Y = {{W/A} \over {\ell /L}}

⇒\Rightarrow Elongation is the same.

Q27
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 ×\times 10−-5 m and density 1.2 ×\times 103 kgm−-3 ? Take viscosity of liquid = 1.8 ×\times 10−-5 Nsm−-2. (Neglect buoyancy due to air).
A 3.8 ×\times 10−-11 N
B 3.9 ×\times 10−-10 N
C 1.8 ×\times 10−-10 N
D 5.8 ×\times 10−-10 N
Correct Answer
Option B
Solution

Viscous force = Weight

=ρ×(43πr3)g= \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g

= 3.9 ×\times 10−-10

Q28
A cylinder of height 2020 mm is completely filled with water. The velocity of efflux of water (in ms−1m{s^{ - 1}}) through a small hole on the side wall of the cylinder near its bottom is
A 1010
B 2020
C 25.525.5
D 55
Correct Answer
Option B
Solution

The velocity of efflux of water is given

v=2ghv = \sqrt {2gh}

Here

hh

is the height of the free surface of water from the hole ∴\therefore

v=2×10×20=20m/sv = \sqrt {2 \times 10 \times 20} = 20m/s
Q29
According to Newton's law of cooling, the rate of cooling of a body is proportional to (Δθ)n,{\left( {\Delta \theta } \right)^n}, where Δθ{\Delta \theta } is the difference of the temperature of the body and the surrounding, and nn is equal to :
A two
B three
C four
D one
Correct Answer
Option D
Solution

Newton's law of cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings.

Mathematically, this law is often written as : dθdt∝Δθ \dfrac{d\theta}{dt} \propto \Delta \theta In this formula, dθdt \dfrac{d\theta}{dt} is the rate of cooling (i.e., the rate of change of temperature with respect to time), and Δθ \Delta \theta is the difference in temperature between the body and the environment.

The law does not specify the temperature difference to an exponential power; it simply states a linear proportionality.

Therefore, the correct answer is : Option D : one.

Q30
A wire fixed at the upper end stretches by length ll by applying a force F.F. The work done in stretching is
A 2Fl2Fl
B FlFl
C F2l{F \over {2l}}
D Fl2{{Fl} \over 2}
Correct Answer
Option D
Solution

Work done by constant force in displacing the object by a distance

ℓ\ell

. = Potential energy stored

=12×= {1 \over 2} \times

Stress ×\times Strain ×\times Volume

=12×FA×lL×AL= {1 \over 2} \times {F \over A} \times {l \over L} \times AL
=12Fl= {1 \over 2}Fl
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