MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
JEE
Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Q.1
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is $U_1$, between 999 nm and 1000 nm is $U_2$and between 1499 nm and 1500 nm is $U_3$.[The Wien constant, $b = 2.88 \times 10^6 nm K$].Then
0%
$U_1 = 0$
0%
$U_3 =0$
0%
$U_1 > U_2$
0%
$U_1 < U_2$
Explanation
Solution:
Wien's law: λ
max
= b/T = (2.88×10⁶ nm·K)/2880K = 1000 nm -- so the spectrum peaks exactly in the 999-1000 nm band.
A blackbody's spectral intensity rises from short wavelengths up to the peak and falls away after it, so the band right at the peak (999-1000 nm, U
2
) carries more radiated energy than the band at 499-500 nm (U
1
), which sits well below the peak in the rising part of the curve.
Q.2
A cylinder of radius R made of a material of thermal conductivity $K_1$ is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity $K_2$ . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is
0%
$K_1+K_2$
0%
$\frac {K_1 K_2}{K_1+ K_2 }$
0%
$(K_1+ 3K_2)/4$
0%
$(3K_1+K_2)/4 $
Explanation
Solution:
Here heat flows lengthwise along both the inner cylinder and the outer shell at once (a parallel arrangement), so the effective conductivity is the cross-sectional-area-weighted average of the two:
K
eff
= (K
1
A
1
+K
2
A
2
)/(A
1
+A
2
).
Inner area A
1
=πR². Outer shell area A
2
=π(2R)²-πR²=3πR².
K
eff
= (K
1
πR²+3K
2
πR²)/(4πR²) = (K
1
+3K
2
)/4.
Q.3
Temperature of the star is determined by
0%
distance
0%
colour
0%
size
0%
None of these
Explanation
Solution:
A star's spectrum -- the mix of colours (wavelengths) it glows in -- shifts with temperature via Wien's law (peak wavelength ∝ 1/T). A star's colour is what astronomers actually read its surface temperature from: hotter stars appear more blue-white, cooler ones more red.
Q.4
Planck's constant has dimensions
0%
$[ML^2 T^{-1} ]$
0%
$[ML T^{-1} ]$
0%
$[ML^2 T^{-2} ]$
0%
$[ML^2 T^1 ]$
Explanation
Solution:
From E=hf, h = E/f. Energy has dimension [ML²T⁻²] and frequency has dimension [T⁻¹] (one over time), so h = [ML²T⁻²] ÷ [T⁻¹] = [ML²T⁻¹]. (This is the same dimension as angular momentum.)
Q.5
Parallel rays of light of intensity $I=912 W/m^2$ are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant $\sigma = 5.7 \times 10^{-8} W^{-2} mK^{-4}$ and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to
0%
330 K
0%
660 K
0%
990 K
0%
1500 K
Explanation
Solution:
At steady state, absorbed power equals emitted power. The sphere only intercepts the parallel rays across its cross-section (πr²), but radiates from its whole surface (4πr²):
I·πr² = σ(T⁴-T
surr
⁴)·4πr² ⇒ I/4 = σ(T⁴-300⁴).
912/4 = 228 = (5.7×10⁻⁸)(T⁴-300⁴) ⇒ T⁴-300⁴ = 4×10ⁿ.
300⁴=8.1×10ⁿ, so T⁴=12.1×10ⁿ, giving T ≈ 330 K.
Q.6
If mass-energy equivalence is taken into account, when water is cooled to from ice, the mass of water should
0%
increase
0%
remain unchanged
0%
decrease
0%
first increase then decrease
Explanation
Solution:
Freezing releases latent heat -- ice at 0°C has LESS internal energy than the same mass of water at 0°C did, since that energy was given off to the surroundings during freezing. By mass-energy equivalence (E=mc²), losing energy corresponds to a tiny decrease in mass, not an increase -- the ice should, strictly, mass very slightly less than the water it formed from.
Q.7
A metal rod of young's modulus Y and coefficient of thermal expansion α is held at its two ends such that its length remains invariant. If its temperature is raised by t°C, the linear stress developed in it is
0%
$ \frac {\alpha t}{Y}$
0%
$\frac {Y}{\alpha t }$
0%
$Y \alpha t$
0%
$\frac {1}{Y \alpha t}$
Explanation
Solution:
If the rod were free, heating it by t°C would produce a strain (a fractional length change) of αt. Since the rod is instead held rigidly at fixed length, that same strain is prevented from occurring -- which means an internal stress builds up to exactly cancel it.
Stress = Young's modulus × strain = Yαt.
Q.8
Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits $10^4$ times the power emitted from B. The ratio $ \frac {\lambda _A}{\lambda _B}$ of their wavelengths at which the peaks occur in their respective radiation curves is
0%
2 :1
0%
1 : 2
0%
3 :1
0%
4 :1
Explanation
Solution:
Total radiated power (Stefan-Boltzmann, for a sphere): P = σ(4πR²)T⁴.
P
A
/P
B
= (R
A
/R
B
)²(T
A
/T
B
)⁴ = (400)²(T
A
/T
B
)⁴ = 10⁴.
(T
A
/T
B
)⁴ = 10⁴/160000 = 1/16, so T
A
/T
B
= 1/2.
Wien's law says λ
max
∝ 1/T, so λ
A
/λ
B
= T
B
/T
A
= 2 -- a ratio of 2:1.
Q.9
A wooden wheel of radius R is made of two semicircular parts(see figure). The two parts are field together by a ring made of a metal strip of cross-sectional area S and length L. L is slightly less than $2 \pi R$. To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta T$ and it just steps over the wheel. As it cools down to the surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$ and its Young's modulus is Y, the force that one part of the wheel applies on the other part is?
0%
$SY \alpha \Delta T$
0%
$ \pi SY \alpha \Delta T$
0%
$2SY \alpha \Delta T$
0%
$ 2 \pi SY \alpha \Delta T$
Explanation
Solution:
Cooling by ΔT, the ring's natural length would shrink by αΔT (a strain), but the rigid wheel prevents that contraction -- inducing a tensile stress YαΔT in the ring, and a ring tension T
ring
= stress × area = SYαΔT.
For a ring under uniform tension, the net force squeezing together everything on one side of any diameter is twice the ring's tension (the same result used for hoop stress in pressure vessels) -- so the force one half of the wheel applies to the other is 2 × SYαΔT = 2SYαΔT.
Q.10
If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be
0%
4
0%
16
0%
32
0%
64
Explanation
Solution:
The power radiated by a sphere scales as R²T⁴ (Stefan-Boltzmann), and since the Earth's distance from the Sun doesn't change, the energy received scales the same way as the total power emitted.
Doubling both R and T multiplies the received energy by 2² × 2⁴ = 4 × 16 = 64.
Q.11
As the temperature is increased, the time period of a pendulum
0%
increases as its effective length increases even though its centre of mass still remains at the centre of the bob.
0%
decreases as its effective length increases even though its centre of mass still remains at the centre of the bob.
0%
increases as its effective length increases due to shifting of centre of mass below the centre of the bob.
0%
decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.
Explanation
Solution:
As temperature rises, the pendulum rod (or wire) undergoes ordinary thermal expansion, making the pendulum's effective length larger, while its centre of mass stays right where it always was, at the centre of the bob. Since T=2π√(L/g), a larger effective length means a longer time period.
Q.12
the increase in moment of inertia I of a uniform rod (coefficient of linear expansion $\alpha$ ) about its perpendicular bisector when its temperature is slightly increased by $\Delta T$ is
0%
$ \frac {I \alpha \Delta}{2}$
0%
$ I \alpha \Delta$
0%
$4 I \alpha \Delta$
0%
$2 I \alpha \Delta$
Explanation
Solution:
I is proportional to L² for a rod about its perpendicular bisector (I=ML²/12), so a small fractional change in length produces roughly TWICE that fractional change in I (from I∝L², ΔI/I ≈ 2ΔL/L).
Thermal expansion gives ΔL/L = αΔT, so ΔI = I × 2αΔT = 2IαΔT.
Q.13
Two conductors having same width and length, thickness d1 and d2 , thermal conductivity k1 and k2 are placed one above the another. Find the equivalent thermal conductivity
0%
$ \frac { (d_1 + d_2)(k_1 d_2 + k_2 d_1)}{2(k_1 + k_2)}$
0%
$ \frac { (d_1 - d_2)(k_1 d_2 + k_2 d_1)}{2(k_1 + k_2)}$
0%
$ \frac { ((k_1 d_2 + k_2 d_1)}{(d_1 + d_2)}$
0%
$ \frac { ((k_1 + k_2 )}{(d_1 + d_2)}$
Explanation
Solution:
For two conducting slabs sharing the same width and length, with thicknesses d
1
,d
2
and conductivities k
1
,k
2
, the effective conductivity of the combination works out to a thickness-weighted average of the two conductivities, combining d
1
with k
1
and d
2
with k
2
in proportion to how much of the total thickness each layer contributes: K
eff
= (k
1
d
1
+k
2
d
2
)/(d
1
+d
2
) is the general form this kind of combined-conductivity result takes.
Q.14
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: The melting point of ice decreases with increase of pressure. Reason: Ice contracts on melting.
0%
(a)
0%
(b)
0%
(c)
0%
(d)
0%
(e)
Explanation
Solution:
Assertion: true -- water is one of the few substances where the solid (ice) is LESS dense than the liquid, and by the Clausius-Clapeyron relation (dP/dT = L/(TΔV)), a substance that contracts on melting (ΔV negative) has a melting point that decreases as pressure increases.
Reason: also true -- ice does contract (i.e. water is denser than ice) when it melts, and this is in fact exactly the property that, through the same Clausius-Clapeyron relation, causes the assertion's pressure effect -- so the reason does correctly explain the assertion here.
Q.15
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: A tube light emits white light. Reason: Emission of light in a tube takes place at a very high temperature
0%
(a)
0%
(b)
0%
(c)
0%
(d)
0%
(e)
Explanation
Solution:
Assertion: true -- a fluorescent tube light does produce white light. Reason: false -- tube lights don't work by heating something to a high temperature the way an incandescent filament does; they work by an electrical discharge exciting mercury vapour to emit ultraviolet light, which then strikes a phosphor coating that fluoresces into visible white light. It's a "cold" light-emission process, not a high-temperature thermal one.
Q.16
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: At room temperature water does not sublimate from ice to steam. Reason: The critical point of water is much above the room temperature.
0%
(a)
0%
(b)
0%
(c)
0%
(d)
0%
(e)
Explanation
Solution:
Assertion: true -- under everyday room conditions, ice (when present) simply melts to liquid water rather than jumping straight to steam.
Reason: true, and it explains why -- water's critical point (about 374°C) is far above room temperature, so at ordinary temperatures water still has a clear liquid-to-gas phase boundary to cross (via boiling), rather than the smooth, boundary-free transition that only becomes possible above the critical point.
Q.17
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: Temperatures near the sea coast are moderate. Reason: Water has high thermal conductivity
0%
(a)
0%
(b)
0%
(c)
0%
(d)
0%
(e)
Explanation
Solution:
Assertion: true -- coastal regions do experience more moderate, less extreme temperature swings than inland areas.
The commonly cited reason for that is water's high SPECIFIC HEAT CAPACITY -- it takes a large amount of energy to change the ocean's temperature, so it heats up and cools down slowly, moderating the nearby coastal climate. Thermal conductivity (how fast heat moves through a material) is a different property and isn't what's driving this effect, so while both statements can be read as true on their own, the reason given isn't the actual explanation for the assertion.
Q.18
The top of an insulated cylindrical container is covered by a disc having radiation emissivity 0.6, thermal conductivity 0.167 $W m^{-1} k^{-1}$ and thickness 1 cm. The temperature is maintained by circulating oil as shown in the figure. The temperature of the upper surface of disc is 127 °C and temperature of the surrounding is 27°C. Neglect the heat loss due to convection. [Given $\sigma\=\frac{17}{3} \tims 10^{-8} Wm^{-2} K^{-4}$ ] Find the rate of radiation loss to the surroundings by unit area of the disc
0%
550 W/m2
0%
595 W/m2
0%
600 W/m2
0%
650 W/m2
Explanation
Solution:
Net radiative loss per unit area = εσ(T⁴-T
surr
⁴), with T=127°C=400 K and T
surr
=27°C=300 K:
400⁴=2.56×10¹⁰, 300⁴=8.1×10⁹, so the difference is 1.75×10¹⁰.
Rate = 0.6 × (17/3×10⁻⁸) × 1.75×10¹⁰ = 595 W/m².
Q.19
Find the temperature of the circulating oil
0%
150 °C
0%
172 ° C
0%
162.6 °C
0%
none of these
Explanation
Solution:
At steady state with no convective loss, whatever heat conducts up through the disc from the oil must equal what radiates away from the top -- 595 W/m² (from the previous part).
Conduction: K(T
oil
-T
upper
)/thickness = 595.
0.167 × (T
oil
-400)/0.01 = 595 ⇒ T
oil
-400 = 595/16.7 ≈ 35.6 ⇒ T
oil
≈ 435.6 K = 162.6°C.
Q.20
Calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 °C and it is defined under which of the following conditions?
0%
From 14.5 °C to 15.5 °C at 760 mm of Hg
0%
From 98.5°C to 99.5 °C at 760 mm of Hg
0%
From 13.5° C to 14.5° C at 76 mm of Hg
0%
From 3.5 ° C to 4.5° C at 76 mm of Hg
Explanation
Solution:
Since water's specific heat varies very slightly with temperature, the calorie needs a precise reference point to be exactly defined -- internationally, it's fixed as the heat needed to raise 1 g of water from 14.5°C to 15.5°C, at standard atmospheric pressure (760 mm of Hg).
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)