Practice Start Test

Properties of Matter

NEET Physics · 97 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is
A 425 kg m −-3
B 800 kg m −-3
C 928 kg m −-3
D 650 kg m −-3
Correct Answer
Option C
Solution

Here,

hoil×ρoil×g=hwater×ρwater×g{h_{oil}} \times {\rho _{oil}} \times g = {h_{water}} \times {\rho _{water}} \times g
ρ0g×140×10−3=ρwg×130×10−3{\rho _0}g \times 140 \times {10^{ - 3}} = {\rho _w}g \times 130 \times {10^{ - 3}}
ρoil=130140×103≈928 kg/m3{\rho _{oil}} = {{130} \over {140}} \times {10^3} \approx 928\,kg/{m^3}
[∵ρw=1 kgm−3]\left[ \because {{\rho _w} = 1\,kg{m^{ - 3}}} \right]
Q32
A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
A 450
B 1000
C 1800
D 225
Correct Answer
Option C
Solution

Given r 1 = 12 cm , r 2 = 6 cm T 1 = 500 K and T 2 = 2 × 500 = 1000 K P 1 = 450 watt Rate of power loss

P∝r2T4P \propto {r^2}{T^4}
P1P2=r12T14r22T24{{{P_1}} \over {{P_2}}} = {{r_1^2T_1^4} \over {r_2^2T_2^4}}
P2=P1r12T14r22T24{P_2} = {P_1}{{r_1^2T_1^4} \over {r_2^2T_2^4}}

Solving we get, P 2 = 1800 watt

Q33
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 o C, while the other one is at 0 o C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is
A 50 o C
B more than 50 o C
C less than 50 o C but greater than 0 o C
D 0 o C
Correct Answer
Option B
Solution

If the final common temperature is T c , C c and C h average heat capacities of cold and hot bodies then as per principle of calorimetry, heat lost = heat gained C h (100°C – T c ) = C c × T c Now T c = C h /(C h + C c ) × 100°C = 100/[1 + (C c /C h )], where C c /C h < 1 It is seen that 1 + C c /C h < 2 Hence, T c > (100/2)°C or T c > 50°C

Q34
Three liquids of densities ρ\rho 1 , ρ\rho 2 and ρ\rho 3 (with ρ\rho 1 > ρ\rho 2 > ρ\rho 3 ), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ\theta 1 , θ\theta 2 and θ\theta 3 obey
A π2>θ1>θ2>θ3≥0{\pi \over 2} > {\theta _1} > {\theta _2} > {\theta _3} \ge 0
B 0 ≤ \le θ\theta 1 < θ\theta 2 < θ\theta 3 < π2{\pi \over 2}
C π2<θ1<θ2<θ3<π{\pi \over 2} < {\theta _1} < {\theta _2} < {\theta _3} < \pi
D π>θ1>θ2>θ3>π2\pi > {\theta _1} > {\theta _2} > {\theta _3} > {\pi \over 2}
Correct Answer
Option B
Solution

Capillary rise,

h=2Tcos⁡θrρgh = {{2T\cos \theta } \over {r\rho g}}

For given value of T and

r,h∝cos⁡θρr,h \propto {{\cos \theta } \over \rho }

Also,

h1=h2=h3{h_1} = {h_2} = {h_3}

or

cos⁡θ1ρ1=cos⁡θ2ρ2=cos⁡θ3ρ3{{\cos {\theta _1}} \over {{\rho _1}}} = {{\cos {\theta _2}} \over {{\rho _2}}} = {{\cos {\theta _3}} \over {{\rho _3}}}

Since,

ρ1>ρ2>ρ3{\rho _1} > {\rho _2} > {\rho _3}

, so

cos⁡θ1>cos⁡θ2>cos⁡θ3{\cos {\theta _1} > \cos {\theta _2} > \cos {\theta _3}}

For

0≤θ<π2,θ1<θ2<θ30 \le \theta < {\pi \over 2},{\theta _1} < {\theta _2} < {\theta _3}

Hence,

0≤θ1<θ2<θ3<π20 \le {\theta _1} < {\theta _2} < {\theta _3} < {\pi \over 2}
Q35
A rectangular film of liquid is extended from (4 cm × \times 2 cm) to (5 cm × \times 4 cm). If the work done is 3 × \times 10 −-4 J, the value of the surface tension of the liquid is
A 0.250 N m −-1
B 0.125 N m −-1
C 0.2 N m −-1
D 8.0 N m −-1
Correct Answer
Option B
Solution

We have d = 2700 m, ρ = 10 3 kg/m3, compressibility = 45.4 × 10 −11 /pascal Now the pressure at bottom of ocean is p = ρgd = 10 3 × 10 × 2700 = 27 × 10 6 Pa Hence, fractional compression = 45.4 × 10 −11 × 27 × 10 6 = 1.2 × 10 −2 = 0.125 N m −-1

Q36
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be
A 74 {7 \over 4}\, T
B 32{3 \over 2} T
C 43{4 \over 3} T
D T
Correct Answer
Option B
Solution

According to Newton’s law of cooling,

dTdt=K(T−Ts){{dT} \over {dt}} = K\left( {T - {T_s}} \right)

For two cases,

dT1dt=K(T1−Ts){{d{T_1}} \over {dt}} = K\left( {{T_1} - {T_s}} \right)

and

dT2dt=K(T2−Ts){{d{T_2}} \over {dt}} = K\left( {{T_2} - {T_s}} \right)

Here,

Ts=T,T1=3T+2T2=2.5T{T_s} = T,{T_1} = {{3T + 2T} \over 2} = 2.5T

and

dT1dt=3T−2T10=T10{{d{T_1}} \over {dt}} = {{3T - 2T} \over {10}} = {T \over {10}}
T2=2T+T′2{T_2} = {{2T + T'} \over 2}

and

dT2dt=2T−T′10{{d{T_2}} \over {dt}} = {{2T - T'} \over {10}}

So

T10=K(2.5T−T){T \over {10}} = K\left( {2.5T - T} \right)

...(i)

2T−T′10=K(2T+T′2−T){{2T - T'} \over {10}} = K\left( {{{2T + T'} \over 2} - T} \right)

...(ii) Dividing eqn. (i) by eqn. (ii), we get

T2T−T′=(2.5T−T)(2T+T′2−T){T \over {2T - T'}} = {{\left( {2.5T - T} \right)} \over {\left( {{{2T + T'} \over 2} - T} \right)}}
2T+T′2−T=(2T−T′)×32{{2T + T'} \over 2} - T = \left( {2T - T'} \right) \times {3 \over 2}
T′=3(2T−T′)⇒4T′=6TT' = 3\left( {2T - T'} \right) \Rightarrow 4T' = 6T
∴T′=32T\therefore T' = {3 \over 2}T
Q37
Two non-mixing liquids of densities ρ\rho and nρ\rho (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length ρ\rho L (ρ\rho < 1) in the denser liquid. The density d is equal to
A {2+(n−1)p}ρ\left\{ {2 + \left( {n - 1} \right)p} \right\}\rho
B {1+(n−1)p}ρ\left\{ {1 + \left( {n - 1} \right)p} \right\}\rho
C {1+(n+1)p}ρ\left\{ {1 + \left( {n + 1} \right)p} \right\}\rho
D {2+(n+1)p}ρ\left\{ {2 + \left( {n + 1} \right)p} \right\}\rho
Correct Answer
Option B
Solution

According to Archimedes principle, Weight of the cylinder = (upthrust) 1 + (upthrust) 2 i.e.,

ALdg=(1−p)LAρg+(pLA)nρgALdg = \left( {1 - p} \right)LA\rho g + \left( {pLA} \right)n\rho g
⇒d=(1−p)ρ+pnρ\Rightarrow d = \left( {1 - p} \right)\rho + pn\rho
=ρ−pρ+npρ= \rho - p\rho + np\rho
=ρ+(n+1)pρ= \rho + \left( {n + 1} \right)p\rho
=p[1+(n−1)p]= p\left[ {1 + (n - 1)p} \right]
Q38
A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U 1 , at wavelength 500 nm is U 2 and that at 1000 nm is U 3 . Wien's constant, b = 2.88 × \times 10 6 nm K. Which of the following is correct ?
A U 1 > U 2
B U 2 > U 1
C U 1 = 0
D U 3 = 0
Correct Answer
Option B
Solution

According to wein's displacement law, maximum amount of emitted radiation corresponding to

λm=bT{\lambda _m} = {b \over T}
λm=2.88×106nmK5760K=500 nm{\lambda _m} = {{2.88 \times {{10}^6}nmK} \over {5760K}} = 500\,nm

From the graph U 1 < U 2 > U 3

Q39
Coefficient of linear expansion of brass and steel rods are α\alpha 1 and α\alpha 2 . Lengths of brass and steel rods are ll 1 and ll 2 respectively. If (ll 1 −- ll 2 ) is maintained same at all temperatures, which one of the following relations holds good?
A α\alpha 1 ll 2 = α\alpha 2 2 ll 1
B α\alpha 1 ll 1 = α\alpha 2 l 2
C α\alpha 1 ll 2 = α\alpha 2 ll 1
D α\alpha 1 ll 2 2 = α\alpha 2 ll 1 2
Correct Answer
Option B
Solution

From question,

(l2−l1)\left( {{l_2} - {l_1}} \right)

is maintained same at all temperatures hence change in length for both rods should be same i.e.,

Δl1=Δl2\Delta {l_1} = \Delta {l_2}

As we know, coefficient of linear expansion,

α=Δll0ΔT\alpha = {{\Delta l} \over {{l_0}\Delta T}}
l1α1ΔT=l2α2ΔT{l_1}{\alpha _1}\Delta T = {l_2}{\alpha _2}\Delta T
l1α1=l2α2{l_1}{\alpha _1} = {l_2}{\alpha _2}
Q40
A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h [Latent heat of ice is 3.4×1053.4 \times {10^5} J/kg and g = 10 N/Kg]
A 136 km
B 68 km
C 34 km
D 544 km
Correct Answer
Option A
Solution

Energy gained by the ice during its fall E = mgh As,

mgh4=mLf{{mgh} \over 4} = m{L_f}
h=mLf×4mg=Lf×4gh = {{m{L_f} \times 4} \over {mg}} = {{{L_f} \times 4} \over g}
=3.4×105×410=136000 m=136 km= {{3.4 \times {{10}^5} \times 4} \over {10}} = 136000\,m = 136\,km
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