Practice Start Test

Properties of Matter

JEE Physics · 167 questions · Page 10 of 17 · Click an option or "Show Solution" to reveal answer

Q91
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 ×\times 103 kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 ×\times 1011 Pa, g = 9.8 m/s2]
A 3.60 ×\times 10−-8
B 2.60 ×\times 10−-7
C 1.87 ×\times 10−-3
D 7.07 ×\times 10−-4
Correct Answer
Option B
Solution

Force on each column =

mg4{{mg} \over 4}

Strain =

mg4AY{{mg} \over {4AY}}
=50×103×9.84×π(1−0.25)×2×1011= {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}

= 2.6 ×\times 10−-7

Q92
A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :
A L1L2\sqrt {{L_1}{L_2}}
B L1+L22{{{L_1} + {L_2}} \over 2}
C 2L1−L22{L_1} - {L_2}
D 3L1−2L23{L_1} - 2{L_2}
Correct Answer
Option C
Solution
y=FLAΔLy = {{FL} \over {A\Delta L}}
⇒ΔL=FLAy\Rightarrow \Delta L = {{FL} \over {Ay}}
⇒L1=L+(1g)LAy\Rightarrow {L_1} = L + {{(1g)L} \over {Ay}}

..... (i) and

L2=L+(2g)LAy{L_2} = L + {{(2g)L} \over {Ay}}

..... (ii)

⇒L=2L1−L2\Rightarrow L = 2{L_1} - {L_2}
Q93
A water drop of radius 1 μ\mum falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 ×\times 10−-5 Nsm−-2 and its density is negligible as compared to that of water 106 gm−-3. Terminal velocity of the water drop is : (Take acceleration due to gravity = 10 ms−-2)
A 145.4 ×\times 10−-6 ms−-1
B 118.0 ×\times 10−-6 ms−-1
C 132.6 ×\times 10−-6 ms−-1
D 123.4 ×\times 10−-6 ms−-1
Correct Answer
Option D
Solution
6πηrv=mg6\pi \eta rv = mg
6πηrv=43πr3ρg6\pi \eta rv = {4 \over 3}\pi {r^3}\rho g

or

v=29ρr2gη=29×103×(10−6)2×101.8×10−5v = {2 \over 9}{{\rho {r^2}g} \over \eta } = {2 \over 9} \times {{{{10}^3} \times {{({{10}^{ - 6}})}^2} \times 10} \over {1.8 \times {{10}^{ - 5}}}}
=123.4×10−6= 123.4 \times {10^{ - 6}}

m/s

Q94
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity. Reason R : Coefficient of viscosity = ForceVelocity gradient{{Force} \over {Velocity\,gradient}} Choose the correct answer from the options given below :
A Both A and R are true, and R is the correct explanation of A.
B Both A and R are true but R is NOT the correct explanation of A.
C A is true but R is false.
D A is false but R is true.
Correct Answer
Option C
Solution

[Pressure][Time] =

[ForceArea]\left[ {{{Force} \over {Area}}} \right]
[distanceArea]\left[ {{{distance} \over {Area}}} \right]

[Coefficient of viscosity] =

[ForceArea]\left[ {{{Force} \over {Area}}} \right]
[distanceArea]\left[ {{{distance} \over {Area}}} \right]

Statement 'A' is true But Statement 'R' is false are coefficient of viscosity

=ForceArea×Velocity gradient= {{Force} \over {Area \times Velocity\,gradient}}
Q95
When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is :
A 19.6 m
B 29.4 m
C 39.2 m
D 73.5 m
Correct Answer
Option B
Solution
t1=2hg{t_1} = \sqrt {{{2h} \over g}}
=2×4.99.8=1 s= \sqrt {{{2 \times 4.9} \over {9.8}}} = 1\,s
Δt=4−1=3 s\Delta t = 4 - 1 = 3\,s

,

v=2gh=2×9.8×4.9=9.8v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 4.9} = 9.8

m/s ∴\therefore depth

=9.8×3=29.4= 9.8 \times 3 = 29.4

m

Q96
The velocity of a small ball of mass 'm' and density d1, when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is d2, then the viscous force acting on the ball, will be :
A mg(1−d1d2)mg\left( {1 - {{{d_1}} \over {{d_2}}}} \right)
B mg(1−d2d1)mg\left( {1 - {{{d_2}} \over {{d_1}}}} \right)
C mg(d1d2−1)mg\left( {{{{d_1}} \over {{d_2}}} - 1} \right)
D mg(d2d1−1)mg\left( {{{{d_2}} \over {{d_1}}} - 1} \right)
Correct Answer
Option B
Solution

Viscous force acting on the ball will be equal and opposite to net of weight and buoyant force

⇒F0=43πr3d1g−43πr3d2g\Rightarrow {F_0} = {4 \over 3}\pi {r^3}{d_1}g - {4 \over 3}\pi {r^3}{d_2}g
=43πr3d1g(1−d2d1)= {4 \over 3}\pi {r^3}{d_1}g\left( {1 - {{{d_2}} \over {{d_1}}}} \right)
=mg(1−d2d1)= mg\left( {1 - {{{d_2}} \over {{d_1}}}} \right)
Q97
If p is the density and η\eta is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is :
A Re=ηdρv{R_e} = {{\eta d} \over {\rho v}}
B Re=ρvηd{R_e} = {{\rho v} \over {\eta d}}
C Re=ρvdη{R_e} = {{\rho vd} \over \eta }
D Re=ηρvd{R_e} = {\eta \over {\rho vd}}
Correct Answer
Option C
Solution

The Reynolds number (Re) is a dimensionless quantity used in fluid mechanics to predict the onset of turbulence.

It is defined as:

Re=ρvdηRe = \frac{{\rho v d}}{{\eta}}

where: ρ\rho is the fluid density vv is the fluid velocity dd is a characteristic linear dimension (for a pipe, this is typically the diameter) η\eta is the dynamic viscosity of the fluid So, Option C is the correct answer:

Re=ρvdη{R_e} = {{\rho vd} \over \eta }
Q98
Potential energy as a function of r is given by U=Ar10−Br5U = {A \over {{r^{10}}}} - {B \over {{r^5}}}, where r is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be :
A (AB)15{\left( {{A \over B}} \right)^{{1 \over 5}}}
B (BA)15{\left( {{B \over A}} \right)^{{1 \over 5}}}
C (2AB)15{\left( {{2A \over B}} \right)^{{1 \over 5}}}
D (B2A)15{\left( {{B \over 2A}} \right)^{{1 \over 5}}}
Correct Answer
Option C
Solution

For equilibrium

−dUdr=0=10Ar11−5Br6- {{dU} \over {dr}} = 0 = {{10A} \over {{r^{11}}}} - {{5B} \over {{r^6}}}
⇒r5=2AB\Rightarrow {r^5} = {{2A} \over B}

And

r=(2AB)1/5r = {\left( {{{2A} \over B}} \right)^{1/5}}
Q99
The bulk modulus of a liquid is 3 ×\times 1010 Nm−-2. The pressure required to reduce the volume of liquid by 2% is :
A 3 ×\times 108 Nm−-2
B 9 ×\times 108 Nm−-2
C 6 ×\times 108 Nm−-2
D 12 ×\times 108 Nm−-2
Correct Answer
Option C
Solution

∵\because

B=ΔP(−ΔVV)B = {{\Delta P} \over {\left( { - {{\Delta V} \over V}} \right)}}
⇒ΔP=3×1010×(0.02)\Rightarrow \Delta P = 3 \times {10^{10}} \times (0.02)
=6×108= 6 \times {10^8}

N/m2

Q100
A drop of liquid of density ρ\rho is floating half immersed in a liquid of density σ{\sigma} and surface tension 7.5×10−47.5 \times 10^{-4} Ncm−-1. The radius of drop in cm\mathrm{cm} will be : (g = 10 ms−-2)
A 15(2ρ−σ) \dfrac{15}{\sqrt{(2 \rho-\sigma)}}
B 15(ρ−σ)\dfrac{15}{\sqrt{(\rho-\sigma)}}
C 32(ρ−σ)\dfrac{3}{2 \sqrt{(\rho-\sigma)}}
D 320(2ρ−σ)\dfrac{3}{20 \sqrt{(2 \rho-\sigma)}}
Correct Answer
Option A
Solution

At equilibrium, forces balance each other

S(2πr)+Fb=mg\mathrm{S}(2 \pi \mathrm{r})+\mathrm{F}_{\mathrm{b}}=m g

Where S=S= surface tension

Fb= buoyant force =23πr3σgS(2πr)=mg−Fb=43πr3(p−σ2)g⇒r2=3S(2p−σ)g⇒r2=3×7.5×10−2(2p−σ)10⇒r2=22.5×10−2(2p−σ)10\begin{aligned} & \mathrm{F}_{\mathrm{b}}=\text{ buoyant force }=\frac{2}{3} \pi r^3 \sigma g \\\\ & S(2 \pi r)=m g-\mathrm{F}_b=\frac{4}{3} \pi r^3\left(p-\frac{\sigma}{2}\right) g \\\\ & \Rightarrow r^2=\frac{3 S}{(2 p-\sigma) g} \\\\ & \Rightarrow r^2 =\frac{3 \times 7.5 \times 10^{-2}}{(2 p-\sigma) 10} \\\\ & \Rightarrow r^2 =\frac{22.5 \times 10^{-2}}{(2 p-\sigma) 10} \end{aligned}

⇒\Rightarrow

r=1.5×10−12p−σmr=\frac{1.5 \times 10^{-1}}{\sqrt{2 p-\sigma}} \mathrm{m}
=152p−σcm=\frac{15}{\sqrt{2 p-\sigma}} \mathrm{cm}
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