Practice Start Test

Properties of Matter

NEET Physics · 97 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
Steam at 100 o C is passed into 20 g of water at 10 o C. When water acquires a temperature of 80 o C, the mass of water present will be [Take specific heat of water = 1 cal g −-1 o C −-1 and latent heat of steam = 540 cal g −-1 ]
A 24 g
B 31.5 g
C 42.5 g
D 22.5 g
Correct Answer
Option D
Solution

According to the principle of calorimetry. Heat lost = Heat gained

mLv+mswΔθ=mwswΔθm{L_v} + m{s_w}\Delta \theta = {m_w}{s_w}\Delta \theta

⇒\Rightarrow m × 540 + m × 1 × (100 – 80) = 20 × 1 × (80 – 10) ⇒\Rightarrow m = 2.5 g Therefore total mass of water at 80°C = (20 + 2.5) g = 22.5 g

Q52
Copper of fixed volume V is drawn into wire of length ll. When this wire is subjected to a constant force F, the extension produced in the wire is Δ\Delta ll. Which of the following graphs is a straight line ?
A Δ\Delta ll versus 1/ll
B Δ\Delta ll versus ll 2
C Δ\Delta ll versus 1/ll 2
D Δ\Delta ll versus ll
Correct Answer
Option B
Solution

From Young Modulus

Y=Fl/AΔl=Fl/(V/l)ΔlY = Fl/A\Delta l = Fl/\left( {V/l} \right)\Delta l

Now

Δl=Fl2/VY\Delta l = F{l^2}/VY

Also

Δl∝l2\Delta l \propto {l^2}
Q53
The density of water at 20 o C is 998 kg/m 3 and at 40 o C is 992 kg/m 3 . The coefficient of volume expansion of water is
A 3 × \times 10 −-4 / o C
B 2 × \times 10 −-4 / o C
C 6 × \times 10 −-4 / o C
D 10 × \times 10 −-4 / o C
Correct Answer
Option A
Solution

From question,

Δρ\Delta \rho

= (998 – 992) kg/m 3 = 6 kg/m 3

ρ=998+9922kg/m3=995 kg/m3\rho = {{998 + 992} \over 2}kg/{m^3} = 995\,kg/{m^3}
ρ=mv\rho = {m \over v}
⇒Δρρ=−ΔVV⇒∣Δρρ∣=∣ΔVV∣\Rightarrow {{\Delta \rho } \over \rho } = - {{\Delta V} \over V} \Rightarrow \left| {{{\Delta \rho } \over \rho }} \right| = \left| {{{\Delta V} \over V}} \right|

∴\therefore Coefficient of volume expansion of water,

1VΔVΔt=1ρΔρΔt=6995×20≈3×10−4/oC{1 \over V}{{\Delta V} \over {\Delta t}} = {1 \over \rho }{{\Delta \rho } \over {\Delta t}} = {6 \over {995 \times 20}} \approx 3 \times {10^{ - 4}}{/^o}C
Q54
Two metal rods 1 and 2 of same lengths have same temperature difference between their ends. Their thermal conductivities are K 1 and K 2 and cross sectional areas A 1 and A 2 , respectively. If the rate of heat conduction in 1 is four times that in 2, then
A K 1 A 1 = 4K 2 A 2
B K 1 A 1 = 2K 2 A 2
C 4K 1 A 1 = K 2 A 2
D K 1 A 1 = K 2 A 2
Correct Answer
Option A
Solution

Let L be length of each rod. Rate of heat flow in rod 1 for the temperature difference

ΔT\Delta T

is

H1=K1A1ΔTL{H_1} = {{{K_1}{A_1}\Delta T} \over L}

Rate of heat flow in rod 2 for the same difference

ΔT\Delta T

is

H2=K2A2ΔTL{H_2} = {{{K_2}{A_2}\Delta T} \over L}

As per question, H 1 = 4H 2

K1A1ΔTL=4K2A2ΔTL{{{K_1}{A_1}\Delta T} \over L} = 4{{{K_2}{A_2}\Delta T} \over L}
⇒K1A1=4K2A2\Rightarrow {K_1}{A_1} = 4{K_2}{A_2}
Q55
If the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are p, q and s respectively, then the corresponding ratio of increase in their lengths would be
A (5q7sp2)\left( {{{5q} \over {7s{p^2}}}} \right)
B (7q5sp2)\left( {{{7q} \over {5s{p^2}}}} \right)
C (2q5sp)\left( {{{2q} \over {5sp}}} \right)
D (7q5sp)\left( {{{7q} \over {5sp}}} \right)
Correct Answer
Option B
Solution

From formula, Increase in length

ΔL=FLAY=4FLπD2Y\Delta L = {{FL} \over {AY}} = {{4FL} \over {\pi {D^2}Y}}
ΔLSΔLC=FSFC(DCDS)2YCYSLSLC{{\Delta {L_S}} \over {\Delta {L_C}}} = {{{F_S}} \over {{F_C}}}{\left( {{{{D_C}} \over {{D_S}}}} \right)^2}{{{Y_C}} \over {{Y_S}}}{{{L_S}} \over {{L_C}}}
=75×(1p)2(1s)q=7q(5sp2)= {7 \over 5} \times {\left( {{1 \over p}} \right)^2}\left( {{1 \over s}} \right)q = {{7q} \over {\left( {5s{p^2}} \right)}}
Q56
A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this which of the following statements is correct?
A The velocity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe.
B Velocity and pressure both are maximum at the narrowest part of the pipe.
C velocity and pressure both are maximum at the widest part of the pipe.
D The velocity is minimum at the narrowest part of the pipe and the pressure is minimum at the widest part of the pipe.
Correct Answer
Option A
Solution

According to Bernoulli’s theorem,

P+12ρv2P + {1 \over 2}\rho {v^2}

= constant and Av = constant If A is minimum, v is maximum, P is minimum.

Q57
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
A length = 200 cm, diameter = 2 mm
B length = 300 cm, diameter = 3 mm
C length = 50 cm, diameter = 0.5 mm
D length = 100 cm, diameter = 1 mm
Correct Answer
Option C
Solution
F=YAL×ℓF = {{YA} \over L} \times \ell

So, extension,

ℓ∝LA∝LD2\ell \propto {L \over A} \propto {L \over {{D^2}}}

[∴\therefore F and Y are constant]

ℓ1∝10012∝100{\ell _1} \propto {{100} \over {{1^2}}} \propto 100

and

ℓ2∝20022∝50{\ell _2} \propto {{200} \over {{2^2}}} \propto 50
ℓ3∝30032∝1003{\ell _3} \propto {{300} \over {{3^2}}} \propto {{100} \over 3}

and

ℓ4∝5014∝200{\ell _4} \propto {{50} \over {{1 \over 4}}} \propto 200

The ratio of

LD2{L \over {{D^2}}}

is maximum for case (length = 50 cm, diameter = 0.5 mm)

Q58
The wettability of a surface by a liquid depends primarily on
A density
B angle of contact between the surface and the liquid
C viscosity
D surface tension
Correct Answer
Option B
Solution

The wettability of a surface by a liquid depends primarily on angle of contact between the surface and the liquid.

Q59
A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using
A Kirchhoff's Law
B Newton's Law of cooling
C Stefan's Law
D Wien's displacement Law
Correct Answer
Option D
Solution

According to Wien’s displacement law

λmT{\lambda _m}T

= constant

λm=constantT{\lambda _m} = {{constant} \over T}

So when a piece of iron is heated,

λm{\lambda _m}

decreases i.e. with rise in temperature the maximum intensity of radiation emitted gets shifted towards the shorter wavelengths.

So the colour of the heated object will change that of longer wavelength (red) to that of shorter (reddish yellow) and when the temperature is sufficiently high and all wavelengths are emitted, the colour will become white.

Q60
The molar specific heats of an ideal gas at constant pressure and volume are denoted by C p and C v , respectively. If γ\gamma = CpCv{{{C_p}} \over {{C_v}}} and R is the universal gas constant, then C v is equal to
A (γ−1)R{{\left( {\gamma - 1} \right)} \over R}
B γR\gamma R
C 1+γ1−γ{{1 + \gamma } \over {1 - \gamma }}
D R(γ−1){R \over {\left( {\gamma - 1} \right)}}
Correct Answer
Option D
Solution

In case of ideal gas C p – C v = R but γ\gamma = C p /C v So,

Cv=Rγ−1{C_v} = {R \over {\gamma - 1}}
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