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Properties of Matter

NEET Physics · 97 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
Water rises to a height h in capillary tube. If the length of capillary tube above the surface of water is made less than h, then
A water rises upto a point a little below the top and stays there.
B water does not rise at all.
C water rises upto the tip of capillary tube and then starts overflowing like a fountain.
D water rises upto the top of capillary tube and stays there without overflowing.
Correct Answer
Option D
Solution

Water will not overflow but will change its radius of curvature.

Q42
The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of
A 4 : 1
B 1 : 1
C 1 : 2
D 2 : 1
Correct Answer
Option D
Solution

Young's modulus

Y=WA.lΔlY = {W \over A}.{l \over {\Delta l}}
W1Y1=W2Y2{{{W_1}} \over {{Y_1}}} = {{{W_2}} \over {{Y_2}}}

[

∵A,l,Δl\because A,l,\Delta l

same for both brass and steel]

W1W2=Y1Y2=2{{{W_1}} \over {{W_2}}} = {{{Y_1}} \over {{Y_2}}} = 2
[YsteelYbrass=2 given]\left[ {{{{Y_{steel}}} \over {{Y_{brass}}}} = 2\,given} \right]
Q43
The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is
A VR2n3r2{{VR{}^2} \over {{n^3}{r^2}}}
B V2Rnr{{{V^2}R} \over {nr}}
C VR2n2r2{{V{R^2}} \over {{n^2}{r^2}}}
D VR2nr2{{V{R^2}} \over {n{r^2}}}
Correct Answer
Option D
Solution

Let the speed of the ejection of the liquid through the holes be v. Then according to the equation of continuity,

πR2V=nπr2v\pi {R^2}V = n\pi {r^2}v
⇒v=πR2Vnπr2=VR2nr2\Rightarrow v = {{\pi {R^2}V} \over {n\pi {r^2}}} = {{V{R^2}} \over {n{r^2}}}
Q44
The value of coefficient of volume expansion of glycerin is 5 × \times 10 −-4 K −-1 . The fractional change in the density of glycerin for a rise of 40 o C in its temperature, is
A 0.025
B 0.010
C 0.015
D 0.020
Correct Answer
Option D
Solution

Let r 0 and r T be densities of glycerin at 0°C and T°C respectively. Then,

ρT=ρ0(1−γΔT){\rho _T} = {\rho _0}\left( {1 - \gamma \Delta T} \right)

where γ\gamma is the coefficient of volume expansion of glycerine and

ΔT{\Delta T}

is rise in temperature.

ρTρ0=1−γΔT⇒γΔT=1−ρTρ0{{{\rho _T}} \over {{\rho _0}}} = 1 - \gamma \Delta T \Rightarrow \gamma \Delta T = 1 - {{{\rho _T}} \over {{\rho _0}}}

Thus,

ρ0−ρTρ0=γΔT{{{\rho _0} - {\rho _T}} \over {{\rho _0}}} = \gamma \Delta T

Here,

γ=5×10−4K−1\gamma = 5 \times {10^{ - 4}}{K^{ - 1}}

and

ΔT=40oC=40K\Delta T = {40^o}C = 40K

∴\therefore The fractional change in the density of glycerin

=ρ0−ρTρ0=γΔT=(5×10−4K−1)(40K)=0.020= {{{\rho _0} - {\rho _T}} \over {{\rho _0}}} = \gamma \Delta T = \left( {5 \times {{10}^{ - 4}}{K^{ - 1}}} \right)\left( {40K} \right) = 0.020
Q45
The two ends of a metal rod are maintained at temperatures 100 o C and 110 o C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200 o C and 210 o C, the rate of heat flow will be
A 8.0 J/s
B 4.0 J/s
C 44.0 J/s
D 16.8 J/s
Correct Answer
Option B
Solution

As the temperature difference

ΔT\Delta T

= 10°C as well as the thermal resistance is same for both the cases, so thermal current or rate of heat flow will also be same for both the cases.

Q46
The approximate depth of an ocean is 2700 m. The compressiblity of water is 45.4 × \times 10 −-11 Pa −-1 and density of water is 10 3 kg/m 3 . What fractional compression of water will be obtained at the bottom of the ocean?
A 1.2 × \times 10 −-2
B 1.4 × \times 10 −-2
C 0.8 × \times 10 −-2
D 1.0 × \times 10 −-2
Correct Answer
Option A
Solution

Compressibility of water, K = 45.4 × 10 –11 Pa –1 density of water P = 10 3 kg/m 3 depth of ocean, h = 2700 m We have to find

ΔVV=?{{\Delta V} \over V} = ?

As we know, compressibility,

K=1B=(ΔV/V)P(P=ρgh)K = {1 \over B} = {{\left( {\Delta V/V} \right)} \over P}\left( {P = \rho gh} \right)

So,

(ΔVV)=Kρgh\left( {{{\Delta V} \over V}} \right) = K\rho gh

= 45.4 × 10 –11 × 10 3 × 10 × 2700 = 1.2258 × 10 –2

Q47
On observing light from three different starts P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If T P , T Q and T R are the respective absolute temperatures of P, Q and R, then it can be conclued from the above observations that
A T P < T R < T Q
B T P < T Q < T R
C T P > T Q > T R
D T P > T R > T Q
Correct Answer
Option D
Solution

According to Wein’s displacement law

λmT{\lambda _m}T

= constant …(i) For star P, intensity of violet colour is maximum.

For star Q, intensity of red colour is maximum.

For star R, intensity of green colour is maximum.

Also,

λr>λg>λv{\lambda _r} > {\lambda _g} > {\lambda _v}

Using equation (i), T r < T g < T v T Q < T R < T P

Q48
A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m 2 . Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (ρair=1.2kg/m3)({\rho _{air}} = 1.2kg/{m^3})
A 2.4 × \times 10 5 N, upwards
B 2.4 × \times 10 5 N, downwards
C 4.8 × \times 10 5 N, downwards
D 4.8 × \times 10 5 N, upwards
Correct Answer
Option A
Solution

Using Bernoulli's theorem and assuming density constant, P 1 +1/2ρv 1 2 = P 2 +1/2ρv 2 2 where: P 2 = pressure outside house P 1 = pressure inside house v 1 = speed of air inside house v 2 = speed of air outside house Pressure difference, P 1 – P 2 = 1/2ρ [v 2 2 – v 1 2 ] Now, P 1 – P 2 = 1/2 × 1.2 [40 2 – 0 2 ] P 1 – P 2 = 960 N/m 2 Since Pressure P = Force/Area, so force acting on roof will be : F = P × A F = 960 × 250 F = 960 × 1000/4 F = 24×10 4 or 2.4×10 5 N which is acting upward.

Q49
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then
A energy = 4VT (1r−1R)\left( {{1 \over r} - {1 \over R}} \right) is released.
B energy = 3VT(1r+1R)\left( {{1 \over r} + {1 \over R}} \right) is absorbed.
C energy = 3VT(1r−1R)\left( {{1 \over r} - {1 \over R}} \right) is released
D energy is neither released nor absorbed.
Correct Answer
Option C
Solution

As surface area decreases so energy is released. Energy released =

4πR2T[n1/3−1]4\pi {R^2}T\left[ {{n^{1/3}} - 1} \right]

where R = n 1/3 r

=4πR3T[1r−1R]=3VT[1r−1R]= 4\pi {R^3}T\left[ {{1 \over r} - {1 \over R}} \right] = 3VT\left[ {{1 \over r} - {1 \over R}} \right]
Q50
Certain quantity of water cools from 70 o C to 60 o C in the first 5 minutes and to 54 o C in the next 5 minutes. The temperature of the surroundings is
A 45 o C
B 20 o C
C 42 o C
D 10 o C
Correct Answer
Option A
Solution

Let the temperature of surroundings be

θ0{\theta _0}

By Newton's law of cooling

θ1−θ2t=k[θ1−θ22−θ0]{{{\theta _1} - {\theta _2}} \over t} = k\left[ {{{{\theta _1} - {\theta _2}} \over 2} - {\theta _0}} \right]
⇒70−605=k[70+602−θ0]\Rightarrow {{70 - 60} \over 5} = k\left[ {{{70 + 60} \over 2} - {\theta _0}} \right]
⇒2=k[65−θ0]\Rightarrow 2 = k\left[ {65 - {\theta _0}} \right]

Similarly,

60−545=k[60+542−θ0]{{60 - 54} \over 5} = k\left[ {{{60 + 54} \over 2} - {\theta _0}} \right]
⇒65=k[57−θ0]\Rightarrow {6 \over 5} = k\left[ {57 - {\theta _0}} \right]

By dividing (i) by (ii) we have

106=65−θ057−θ0⇒θ0=45o{{10} \over 6} = {{65 - {\theta _0}} \over {57 - {\theta _0}}} \Rightarrow {\theta _0} = {45^o}
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