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CBSE Questions for Class 11 Engineering Physics Thermal Properties Of Matter Quiz 14 - MCQExams.com
CBSE
Class 11 Engineering Physics
Thermal Properties Of Matter
Quiz 14
A 20gm bullet whose specific heat is 5000 J/ (kg- $$^o C$$) and moving at 2000 m/s plunges into a
1.0 kg
block of wax whose specific heat is
3000 J/(kg-
$$^o C$$). Both bullet and wax are $$25^o C$$
and assume that (i) the bullet comes to rest in the wax and (ii) all its kinetic energy goes into heating the wax. Thermal temperature of the wax in $$^o C$$ is close to
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0%
28.1
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31.5
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37.9
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42.1
Two spheres of the same material havethe same radius but one is hollow while the other is solid. Both spheres are heated to same temperature. Then
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The solid sphere expands more
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The hollow sphere expands more
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Expansion is same for both
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Nothing can be about their relative expansion
A vertical column 50$$\mathrm { cm }$$ long at $$50 ^ { \circ } \mathrm { C }$$ balances another column of same liquid 80 an long at 100$$^ { 0 } C$$ The coefficient of absolute expansion of the liquid is
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0.005$$/ ^ { \circ } C$$
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$$0.03/ ^ { 0 } \mathrm { c }$$
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0.002$$/ ^ { 0 } \mathrm { C }$$
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$$0.0002 / ^ { \circ } C$$
Calorimeter helps us to determine
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boiling point
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melting point
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specific heat
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freezing point
If temperature difference between two sides of an iron plate 2 cm thick is $${ 10 }^{ 0 }C$$ and thermal conductivity of iron is $$0.02 K cal/smK$$ then heat transmitted through the plate per second per $${ m }^{ 2 }$$ at steady state is:
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$$100Kcal/{ m }^{ 2 }s$$
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$$10Kcal/{ m }^{ 2 }s$$
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$$600Kcal/{ m }^{ 2 }s$$
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$$40Kcal/{ m }^{ 2 }s$$
What equivalent of calorimeter is :
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$$ mC_1 $$
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$$ \frac {mC_1 }{C_2} $$
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$$ \frac {mC_2 }{C_2} $$
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none of these
Two 50 g ice cubes are placed into 0.2 kg of water in a closed container made of styrofoam. the water is initially a temperature of $$ -15^0 C $$ . the system is allowed to attain equilibrium. the system is allowed to attain equilibrium . the system does not exchange heat with with surrounding . the average specific heat for for ice between $$ -15^0 c $$ and $$ 0^0 C $$ id $$ 2 kJkg^{-1}K^{-1} $$ latent heat of fusion for ice is is $$ \frac {1000}{3} kjkg^{-1} $$ specific heat for water is $$ 4.2 kJkg^{-1}k^{-1} $$
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The amount of heat supplied by water to melt ice is 21 kJ
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The final temperature of the mixture is $$ -12.4^0 C $$
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the final temperature of the system is $$0^0 C $$
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the amount of ice melted is 0.054 kg
Two liquids $$A$$ and $$B$$ are at $$30 ^ { \circ } \mathrm { C }$$ and $$20 ^ { \circ } \mathrm { C }$$ respectively. When they are mixed in equal masses the temperature of the mixture is found to be $$26 ^ { \circ } \mathrm { C }$$ . The ratio of specific heats is
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$$4:3$$
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$$3:4$$
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$$2:3$$
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$$3:2$$
Select the correct statement related to heat
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Heat is possessed by a body
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Hot water contains more heat as compared to cold water
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Heat is a energy which flows due to temperature difference
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All of these
An electric heater rated as 2 kW is used to heat 200 kg of water from $$1{ 0 }^{ \circ }C$$ to $$7{ 0 }^{ \circ }C$$. Assuming no heat losses, the time take is-
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$$25.2\ sec$$
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$$6\times { 10 }^{ 3 }\ sec$$
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$$25.2\times { 10 }^{ 3 }\ sec$$
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$$25.2\times { 10 }^{ 6 }\ sec$$
One mole of a perfect gas in a cylinder fitted with a piston has a pressure P, volume V and temperature is increased by 1 K keeping pressure constant , the increase in volume is
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$$\dfrac { 2V }{ 273 } $$
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$$\dfrac { 2V }{ 91 }$$
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$$\dfrac { V }{ 273 } $$
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V
On a hypothetical scale $$A$$ the ice point is $$42 ^ { \circ }$$ and the steam points is $$182 ^ { \circ }$$ For another scale $$B,$$ the ice point is $$- 10 ^ { \circ }$$ and steam point is $$90 ^ { \circ } $$
If $$B$$ reads $$60 ^ { \circ } ,$$ the reading of $$A$$ is,
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$$160 ^ { \circ }$$
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$$140 ^ { \circ }$$
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$$120 ^ { \circ }$$
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$$110 ^ { \circ }$$
Two identical rods of a metal are welded as shown in figure (a). A certain amount of heat flows through them in 12 min.
If the rods are welded as shown in figure (b) then
the same amount of heat will flow in
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1 minute
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3 minutes
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16 minutes
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2 minutes
The figure shows the face and interface temperature of a composite slab containing four layers of two materials having an identical thickness. Under the steady slate conditions, find the value of temperature $$\theta$$.
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$$5^o$$C
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$$10^o$$C
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$$-15^o$$C
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$$15^o$$C
50 gm of copper is heated to increase its temperature by $${ 10 }^{ \circ }C$$ . If the same quantity of heat is given to 10gm of water, the rise in its temperature is ( Specific heat of copper = $$420\quad Joule-{ kg }^{ -1 }{ C }^{ -1 }$$ )
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$${ 5 }^{ \circ }C$$
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$${ 6 }^{ \circ }C$$
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$${ 7 }^{ \circ }C$$
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$${ 8 }^{ \circ }C$$
Three slabs of equal area and thickness are arranged as shown in the figure. Find the value of $$T_1$$ and $$T_2$$ in steady state:
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$$68^oC$$&$$52^oC$$
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$$62^oC$$&$$58^oC$$
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$$60^oC$$&$$50^oC$$
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$$50^oC$$&$$30^oC$$
Two rods of the same length and cross-sectional area are joined in series. The thermal conductivity of the rods is in ratio 1:The ends are maintained at a temperature T$$_{1}$$ and T$$_{2}$$ as shown in the figure. The sides are thermally insulated. If T$$_{1}$$ > T$$_{2}$$then which of the following graphs represents variation temperature gradient $$ \left( \frac { d T } { dx } \right) $$ with x in steady state?
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0%
0%
0%
An ideal gas with pressure $$P$$, volume $$V$$ and temperature $$T$$ is expanded isothermally to a volume $$2V$$ and a final pressure $${P}_{I}$$. The same gas is expanded adiabatically to a volume $$2V$$, the final pressure is $${P}_{A}$$. In terms of the ratio of the two specific heats for the gas $$\gamma$$, the ratio $${P}_{I}/{P}_{A}$$ is:
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$${2}^{\gamma-1}$$
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$${2}^{1-\gamma}$$
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$${2}^{\gamma}$$
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$$2\gamma$$
The value of the gas constant (R) calculated from the perfect gas equation is $$8.32 \,Joule/gm \,mol \,K$$, whereas its value calculated from the knowledge of $$C_p$$ and $$C_v$$ of the gas is $$1.98 \,cal/gm \,mole \,K$$. from this data, the value of $$J$$ is
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$$4.16 \,J/cal$$
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$$4.18 \,J/cal$$
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$$4.20 \,J/cal$$
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$$4.22 \,J/cal$$
A solid whose volume does not change with temperature floats in a liquid. For two different temperatures $$ t_{1} $$ and$$ t_{2} $$ of the liquid, fractions $$ f_{1} $$ and $$ f_{2} $$ of the volume of the solid remain submerged in the liquid. The coefficient of v
olume expansion of the liquid is equal to
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$$
\frac{f_{1}-f_{2}}{f_{2} t_{1}-f_{1} t_{2}}
$$
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$$
\frac{f_{1}-f_{2}}{f_{1} t_{1}-f_{2} t_{2}}
$$
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$$ \frac{f_{1}+f_{2}}{f_{2} t_{1}+f_{1} t_{2}} $$
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$$
\frac{f_{1}+f_{2}}{f_{1} t_{1}+f_{2} t_{2}}
$$
When the temperature difference between in side and outside of a room is $$20 $$ the rate of heat flow through a window is 273 J/s. If the temperature difference becomes 20K, the rate of flow of heat
through the same window will be:
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253 J/s
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273 J/s
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293 J/s
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None of these
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $${39^\circ }\,W$$ and $${239^\circ }\,W$$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $${39^\circ }$$ on the Celsius scale?
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$${139^\circ }W$$
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$${78^\circ }W$$
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$${117^\circ }W$$
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$${200^\circ }W$$
For $$N_2$$the value of $$C_p-C_v$$=
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$$14R$$
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$$\frac {R}{14}$$
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$$28 R$$
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$$\frac {R}{28}$$
With rise in temperature, density of a given body change according to one of the following relations
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$$\rho = \rho_0 [1+\gamma d \theta]$$
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$$\rho = \rho_0 [1-\gamma d \theta]$$
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$$\rho = \rho_0 \gamma d \theta$$
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$$\rho = \rho_0 /\gamma d \theta$$
Critical temperature of a gas obeying van der waals' equation is :
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$$\dfrac{8a}{27Rb}$$
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$$\dfrac{a}{27b^2r}$$
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$$3b$$
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$$\dfrac{1}{27Rb}$$
One mole of gas ''A'' and 2 moles of S$${ O }_{ 2 }$$ are placed in a container,gaseous can be:
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$${ H }_{ 2 }$$
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$$C{ H }_{ 4 }$$
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He
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$${ D }_{ 2 }$$
Two materials having coefficients of thermal conductivity $$'3K'$$ and $$'K'$$ and thickness 'd' and $$'3d'$$, respectively, are joined to form a slab as shown in the figure. The temperature of the outer surfaces are $$'\theta_2'$$ and $$'\theta_1'$$ respectively, $$(\theta_2 > \theta_1)$$. The temperature at the interface is?
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$$\dfrac{\theta_2+\theta_1}{2}$$
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$$\dfrac{\theta_1}{10}+\dfrac{9\theta_2}{10}$$
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$$\dfrac{\theta_1}{3}+\dfrac{2\theta_2}{3}$$
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$$\dfrac{\theta_1}{6}+\dfrac{5\theta_2}{6}$$
Explanation
Let the temperature of interface be $$"\theta"$$
$$i_1=i_2$$$$\{$$ Steady state conduction$$\}$$
$$\dfrac{ 3KA(\theta_2-\theta)}{d}=\dfrac{KA(\theta -\theta_1)}{3d}$$
$$9\theta_2 - 9\theta =\theta - \theta_1$$
$$9\theta_2+\theta_1=\theta +9\theta$$
$$\theta =\dfrac{9\theta_2}{10}+\dfrac{\theta_1}{10}$$.
The figure shows a system of two thin concentric spherical shells of radii $${r_1}\;{\text{and}}\;{r_2}$$
and kept at temperatures
$${T_1}\;{\text{and}}\;{T_2}$$ respectively. The radial rate of heat in a substance between the two concentric spheres is proportional to :
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$$\dfrac{{{r_1}{r_2}}}{{\left( {{r_2} - {r_1}} \right)}}$$
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$${\left( {{r_2} - {r_1}} \right)}$$
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$$\dfrac{{\left( {{r_2} - {r_1}} \right)}}{{{r_1}{r_2}}}$$
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$$\log \left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)$$
When water is heated from $$0^0$$C to $$10^0$$C, its volume -
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increases
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decreases
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does not change
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first decreases and then increases
In a Carnot engine, when heat is absorbed from the source, temperature of source
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Increases
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Decreases
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Remains constant
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Cannot say
At ordinary temperature,the molecule of an ideal gas have only translation and rotational kinetic energies. At high temperature,they may have vibrational energy.As a result of this, at higher temperature molar specific heat capacity at constant volume $$ C_v $$ for a diatomic gas is
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$$ = \frac {5R}{2} $$
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$$ < \frac {5R}{2} $$
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$$ > \frac {3R}{2} $$
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$$ = \frac {3R}{2} $$
A compound rod is formed of a steel core of diameter $$1cm$$ and outer casing is of copper, whose outer diameter is $$2 cm$$. The length of this compound rod is $$2m$$ and one end is maintained as $$100^{\circ}C$$, and the end is at $$0^{\circ}C$$. If the outer surface of the rod is thermally insulated, then heat current in the rod will be:
(Given : Thermal conductivity of steel $$= 12\ cal/m-K-s$$, thermal conductivity of copper $$= 92\ cal/m-K-s$$).
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$$2\ cal/s$$
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$$1.13\ cal/s$$
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$$1.42\ cal/s$$
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$$2.68\ cal/s$$
If the thermal conductivity of the material of the rod of length $$l$$, is $$K$$, then the rate of heat flow through a tapering rod, tapering from radius $$r_{1}$$ and $$r_{2}$$, if the temperature of the ends are maintained at $$T_{1}$$ and $$T_{2}$$, is
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$$\dfrac {\pi Kr_{1}r_{2}(T_{1} + T_{2})}{l}$$
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$$\dfrac {\pi Kr_{1}r_{2}(T_{1} - T_{2})}{l}$$
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$$\dfrac {\pi Kr_{1}r_{2}(T_{1} - T_{2})}{2l}$$
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$$\dfrac {\pi Kr_{1}^{2}(T_{1} + T_{2})}{l}$$
Five wires each of cross-sectional area A and length l are combined as shown in figure. The thermal conductivity of copper and steel are K1 and K2 respectively. If $$A$$ is maintained at $$100^{\circ}C$$ and $$C$$ is maintained at $$0^{\circ}C$$. The rate emitted at $$C$$ is : (assume the curve surfaces of rods are thermally insulated).
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$$\dfrac {50A(K_{1} + K_{2})}{l}$$
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$$\dfrac {100A(K_{1} + K_{2})}{l}$$
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$$\dfrac {A(K_{1} + K_{2})}{l}$$
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None of these
Which of the following does not affect the distribution of insolation?
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Altitude
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Angle of sun's rays
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Length of the day
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Ocean currents
A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equals to :
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$$F/(3 \pi r^2 YT)$$
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$$3F/( \pi r^2 YT)$$
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$$6F/( \pi r^2 YT)$$
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$$9F/( \pi r^2 YT)$$
Explanation
$$\because $$ Length of cylinder remains unchanged
so $$\left(\dfrac{F}{A}\right)_{Compressive} = \left(\dfrac{F}{A}\right)_{Thermal}$$
$$\dfrac{F}{\pi r^2} = Y \alpha T$$
($$\alpha$$ is linear coefficient of expansion)
$$\therefore \alpha = \dfrac{F}{YT \pi r^2}$$
$$\therefore $$ The coefficient of volume expansion $$\gamma = 3 \alpha$$
$$\therefore \gamma = 3 \dfrac{F}{Y T \pi r^2}$$
The ice is filled in a hollow glass sphere of thickness $$2\ mm$$ and external radius $$10\ cm$$. This hollow glass sphere with ice now placed in a bath containing boiling water at $$100^{\circ}C$$. The rate at which ice melts, is: (Neglect volume change in ice)
(Given : thermal conductivity of glass $$1.1\ W/m/K$$, latent heat of ice $$= 336\times 10^{3} J/kg$$).
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$$\dfrac {m}{t} = 0.01\ kg/s$$
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$$\dfrac {m}{t} = 0.002\ kg/s$$
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$$\dfrac {m}{t} = 0.02\ kg/s$$
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$$\dfrac {m}{t} = 0.001\ kg/s$$
A block of ice at $$0^{\circ}$$ rests on the upper surface of the slab of stone of area $$3600\ cm^{2}$$ and thickness of $$10\ cm$$. The slab is exposed on the lower surface to steam at $$100^{\circ}C$$. If $$4800\ g$$ of ice is melted in one hour, then the thermal conductivity of stone is:
(Given: The latent heat of fusion of ice $$= 80\ cal/gm)$$.
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$$K = 2.96\times 10^{-3} cal/ cm\ s^{\circ}C$$
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$$K = 1.96\times 10^{-3} cal/ cm\ s^{\circ}C$$
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$$K = 0.96\times 10^{-3} cal/ cm\ s^{\circ}C$$
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None of the above
The value of $$ C_p - C_v $$ is 1:00 R for a gas sample in state A and is 1.08 R in state B. Let $$ P_A, P_a $$ denote the pressures and $$ T_A\ and\ T_a $$ denote the temperature of the states $$A$$ and $$a$$ respectively.Most likely
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$$ P_A < P_a\ and\ T_A > T_e $$
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$$ P_A > P_a\ and\ T_A < T_a $$
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$$ P_A = P_a\ and\ T_A < T_a $$
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$$ p_A > P_a\ and\ T_A = T_e $$
One mole of an ideal gas passes through a process where pressure and volume obey the relation $$P = P_0 \left[1 - \dfrac{1}{2} \left(\dfrac{V_0}{V} \right)^2 \right]$$. Here $$P_0$$ and $$V_0$$ are constants . Calculate the change in the temperature of the gas if its volume change from $$V_o $$ to $$2 V_o$$.
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$$\dfrac{1}{2} \dfrac{P_o V_o}{R}$$
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$$\dfrac{3}{4} \dfrac{P_o V_o}{R}$$
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$$ \dfrac{5}{4} \dfrac{P_o V_o}{R}$$
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$$\dfrac{1}{4} \dfrac{P_oV_o}{R}$$
A heater of constant power is used to boil water in a pot. During heating, temperature of water increases from $$ 60^o C$$ to $$65^o C $$ in $$1.0 \ min$$. When the heater is switched off, temperature of water falls from $$ 65^o C$$ to $$60^o C $$ in $$9.0 \ min$$. If rate of heat dissipation to the surrounding remains almost constant, what proportion of heat received by water during heating would be dissipated to the surroundings?
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$$9\%$$
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$$10\%$$
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$$12.5\%$$
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$$15\%$$
After the steady state conditions are reached, the temperature of the spherical end of the rod, $$T_{S}$$ is:
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$$T_{S}=T-\dfrac{P(L+r)}{KA}$$
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$$T_{S}=0^{o}C$$
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$$T_{S}=\dfrac{PL}{KA}$$
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$$T_{S}=T-\dfrac{PL}{KA}$$
Explanation
The distance from the end A of the copper rod (where power is supplied) and centre O of the metal sphere is (L+r). Hence , in the steady state condions,
$$P$$=$$\dfrac{KA(T−T_{S})}{L+r}$$
Which gives
$$T_{S}=T-\dfrac{P(L+r)}{KA}$$
(The heat is trainsmitted up to centre of the spherical end and the sphere loses energy by radiation out of its spherical surface.)
If the metal sphere attached at the end of the copper rod is made of brass, whose thermal conductivity is $$K_{b}<K$$, then which of the following statements is true?
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The temperature of the sphere will, under steady state conditions, continue to be $$T_{B}$$
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The power that will be radiated out from the sphere will still be $$P_{S}$$
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It will take smaller time for steady state conditions to be reaches
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The rate of thermal energy transmitted across the copper rod, under steady state, will be reduced
An alternative method for keeping the tea hot would be to place the teapot on a $$10$$ pound block that has been heated in an oven to $$300^{o}C$$. A block of which of the following substance would best be able to keep the tea hot?
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copper (specific heat $$=0.39\ J/g\ K$$)
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granite (specific heat $$=0.79\ J /g\ K$$)
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iron (specific heat $$=0.45\ J/g\ K$$)
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pewter (specific heat $$=0.17\ J/g\ K$$)
The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. If the atmosphere pressure is $$Hm$$ of mercury and the density of mercury is $$n$$ times that of lake water the depth of lake is :
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$$H/n$$
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$$nH/2$$
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$$nH$$
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$$2nH$$
What is the final temperature of the system ?
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$$48^{o}C$$
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$$72^{o}C$$
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$$94^{o}C$$
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$$100^{o}C$$
The net power that will be radiated out, $$P_{S}$$, from the sphere after steady state conditions are reached is:
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$$P_{S}=P$$
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$$P_{S}=\dfrac{PA}{4\pi r^{2}}$$
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$$P_{S}=0$$
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$$P_{S}=\sigma \varepsilon T_{5}^{4}$$
Explanation
Under steady state conditions, all temperature remain constant including $$T_{S}$$, the temperature at centre of sphere. Hence no thermal energy accumulates inside the sphere. Hence $$P_{S}$$=$$P$$= energy received from source.
If, after some of the tea has been transferred to the thermos (as described in the passage), the teapot with its contents (at a temperature of $$80^{o}C$$) was placed or the insulated warmer for $$5$$ minutes, what would be the temperature at the end of this $$5$$ minutes period (Assume that no significant heat transfer occurs with the surrounding):
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$$80.7\ ^{o}C$$
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$$82.5\ ^{o}C$$
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$$83.2\ ^{o}C$$
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$$95.2\ ^{o}C$$
Three rods of identical cross-sectional area and made from the same metal from the sides of an isosceles triangle $$ABC$$ right angled at $$B$$. The points $$A$$ and $$B$$ are maintained at temperatures $$T$$ and $$\sqrt 2 T$$ respectively in the steady state. Assuming that only heat conduction takes place temperature of point $$C$$ will be:
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$$\dfrac {3T}{\sqrt 2 +1}$$
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$$\dfrac {T}{\sqrt 2 +1}$$
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$$\dfrac {T}{\sqrt 3 (\sqrt 2 +1)}$$
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$$\dfrac {T}{\sqrt 2 -1}$$
Four identical rods which have thermally insulated lateral surfaces are joined at point $$A$$. Points $$B, C, D$$ and $$E$$ are connected to larger reservoirs. If heat flows into the junction from point $$B$$ at rate of $$1\ W$$ and from point $$C$$ at $$3\ W$$ inside, flows out from $$D$$ at $$5\ W$$, which relation (s) is /are correct for temperature of these points?
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$$T_A < T_E$$
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$$T_B =T_C$$
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$$T_C > T_D$$
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$$T_B = T_E$$
Explanation
Heat flow
At A
$$\Rightarrow$$x+1+3=5
x=1
Heat in flows from E
$$T_E>T_A$$
$$T_C>T_A>T_D$$
$$T_B−T_A=T_E−T_A$$
$$T_B=T_E$$
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