CBSE 11th Standard Physics Subject Ncert Exemplar 1 Mark Questions With Solution 2021
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CBSE 11th Standard Physics Subject Ncert Exemplar 1 Mark Questions With Solution 2021
11th Standard CBSE
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Reg.No. :
Physics
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Name the device used for measuring the mass of atoms and molecules.
(a)Mass spectrograph.
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The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?
(a)No, given \(\underset { i }{ \Sigma } { F }_{ i }\neq 0\)
The sum of torques about a certain point O
\(\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }=0\)
The sum of torques about any other point O'.
\(\underset { i }{ \Sigma } ({ r }_{ i }-a)\times F_{ i }=\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }-a\times \underset { i }{ \Sigma } { F }_{ i }\)
Here, the second term need not vanish. -
What is the direction of areal velocity of the earth around the sun?
(a)It is normal to the plane containing the earth and the sun as areal velocity. \(\frac { \triangle A }{ \triangle t } =\frac { 1 }{ 2 } r\times v\triangle t\) and directed according to right hand rule.
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Is stress a vector quantity ?
(a)No, because stress is a scalar quantity, not a vector quantity.
Stress = \(\frac { Magnitude \ of \ internal \ reaction \ force }{ Area \ of \ cross-section } \) -
Can a system be heated and its temperature remain constant?
(a)If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.
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Air pressure in a car tyre increases during driving. Explain.
(a)During driving, temperature of the gas increases while its volume remains constant. So, according to Charles' law, at constant V, p \(\propto \) T. Therefore, pressure of gas increases
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If a refrigerator's door is kept open, will the room become cool or hot?Explain.
(a)Here, heat removed is less than the heat supplied and hence the room become hotter.
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When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased." What about Boyle's law in this case?
(a)When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.
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Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters \(1\overset { 0 }{ A } \) and \(2\overset { 0 }{ A } \). The gases may be considered under identical conditions of temperature, pressure and volume.
(a)As, we know, mean free path,
\(\lambda \propto \frac { 1 }{ { d }^{ 2 } } \)
\(Given,\ { d }_{ 1 }=1\overset { 0 }{ A } \ and\ { d }_{ 2 }=2\overset { 0 }{ A } \Rightarrow { \lambda }_{ 1 }:{ \lambda }_{ 2 }=4:1\) -
What is the quantity transmitted with propagation of longitudinal waves through a medium?
(a)Propagation of longitudinal waves through a medium leads to transmission of energy through the medium.
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How speed of sound waves in air varies with humidity?
(a)Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.
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The displacement of an elastic wave is given by the function y = 3sin \(\omega \)t + 4cos\(\omega \)t Where, y is in cm and t is in second.Calculate the resultant amplitude.
(a)The resultant amplitude will be
\(y=\sqrt { { y }^{ 2 }_{ 1 }+{ y }^{ 2 }_{ 2 } } =\sqrt { 9+16 } =\sqrt { 25 } =5cm\) -
Sound waves of wavelength \(\lambda \) travelling in a medium with a speed of vm/s enter into another medium where its speed is 2vm/s.Wavelength of sound waves in the second medium is
(a)Frequency in the first medium, v =\(\frac { v }{ \lambda } \)
Frequency will remain same in the second medium, as the source.
\(\Rightarrow v^{ , }=v\Rightarrow \frac { 2v }{ \lambda ^{ ' } } =\frac { v }{ \lambda } \Rightarrow { \lambda }^{ ' }=2\lambda \) -
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train begins to move with a speed of 10 ms-1 towards the platform. What is the frequency of the sound for an observer standing on the platform? (Sound velocity in air =330 ms-1).
(a)Given, v=400 Hz, Vs = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V'=apparent frequency
\(=\frac { v\times V }{ v-{ v }_{ s } } =\frac { 330\times 400 }{ 330-10 } \)
\(=412.5\ Hz\) -
A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?
(a)The sonometer frequency is given by
\(v=\frac { n }{ 2L } \sqrt { \frac { T }{ \mu } } \)
Now, as it vibrates with length L, we assume v=v1
n = n1
\(\therefore { v }_{ 1 }=\frac { { n }_{ 1 } }{ 2L } \sqrt { \frac { T }{ \mu } } \)
When length is doubled, then
\({ v }_{ 2 }=\frac { { n }_{ 2 } }{ 2\times 2L } \sqrt { \frac { T }{ \mu } } \)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
To keep the resonance,
\( \frac { { v }_{ 1 } }{ { v }_{ 2 } } =1=\frac { { n }_{ 1 } }{ { n }_{ 2 } } \times 2\)
\(\\ \Rightarrow { n }_{ 2 }=2{ n }_{ 1 }\)
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.
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CBSE 11th Standard Physics Subject Ncert Exemplar 1 Mark Questions With Solution 2021 Answer Keys
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Mass spectrograph.
Mass spectrograph.
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No, given \(\underset { i }{ \Sigma } { F }_{ i }\neq 0\)
The sum of torques about a certain point O
\(\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }=0\)
The sum of torques about any other point O'.
\(\underset { i }{ \Sigma } ({ r }_{ i }-a)\times F_{ i }=\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }-a\times \underset { i }{ \Sigma } { F }_{ i }\)
Here, the second term need not vanish.No, given \(\underset { i }{ \Sigma } { F }_{ i }\neq 0\)
The sum of torques about a certain point O
\(\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }=0\)
The sum of torques about any other point O'.
\(\underset { i }{ \Sigma } ({ r }_{ i }-a)\times F_{ i }=\underset { i }{ \Sigma } { r }_{ i }\times F_{ i }-a\times \underset { i }{ \Sigma } { F }_{ i }\)
Here, the second term need not vanish. -
It is normal to the plane containing the earth and the sun as areal velocity.\(\frac{\Delta A}{\Delta t}=\frac{1}{2} r \times v \Delta t\) and directed according to right hand rule.
It is normal to the plane containing the earth and the sun as areal velocity.\(\frac{\Delta A}{\Delta t}=\frac{1}{2} r \times v \Delta t\) and directed according to right hand rule.
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No, because stress is a scalar quantity, not a vector quantity.
Stress = Magnitude of internal reaction force/Area of cross−sectionNo, because stress is a scalar quantity, not a vector quantity.
Stress = Magnitude of internal reaction force/Area of cross−section -
If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.
If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.
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During driving, temperature of the gas increases while its volume remains constant. So, according to Charles' law, at constant V, p ∝ T. Therefore, pressure of gas increases
During driving, temperature of the gas increases while its volume remains constant. So, according to Charles' law, at constant V, p ∝ T. Therefore, pressure of gas increases
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Here, heat removed is less than the heat supplied and hence the room become hotter.
Here, heat removed is less than the heat supplied and hence the room become hotter.
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When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.
When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.
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As, we know, mean free path,
λ ∝ 1/d2
Given, d1= 1Ao and d2 = 2Ao⇒ λ1 : λ2=4:1As, we know, mean free path,
λ ∝ 1/d2
Given, d1= 1Ao and d2 = 2Ao⇒ λ1 : λ2=4:1 -
Propagation of longitudinal waves through a medium leads to transmission of energy through the medium.
Propagation of longitudinal waves through a medium leads to transmission of energy through the medium.
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Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.
Speed of sound waves in air increases with increases with increase in humidity. This is because presence of moisture decreases the density of air.
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The resultant amplitude will be
y=\(\sqrt{y_2^1+y_2^2}=\sqrt{9+16}=\sqrt{25}=\)5cmThe resultant amplitude will be
y=\(\sqrt{y_2^1+y_2^2}=\sqrt{9+16}=\sqrt{25}=\)5cm -
Frequency in the first medium, v =v/λ
Frequency will remain same in the second medium, as the source.
⇒v, =v ⇒ 2v/λ′ = v/λ ⇒ λ′ = 2λFrequency in the first medium, v =v/λ
Frequency will remain same in the second medium, as the source.
⇒v, =v ⇒ 2v/λ′ = v/λ ⇒ λ′ = 2λ -
Given, v=400 Hz, Vs = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V'=apparent frequency
\(=\frac{v×V}{v−v_s}=\frac{330×400}{330−10}\)
=412.5 HzGiven, v=400 Hz, Vs = 10m/s, v = 330m/s
As the source is moving towards stationary observer, apparent frequency will be more than the original.
V'=apparent frequency
\(=\frac{v×V}{v−v_s}=\frac{330×400}{330−10}\)
=412.5 Hz -
The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v1
n = n1
∴v1=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v2=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n2=2n1
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.The sonometer frequency is given by
v=\(\frac{n}{2L}\sqrt{\frac{T}{μ}}\)
Now, as it vibrates with length L, we assume v=v1
n = n1
∴v1=\(\frac{n_1}{2L}\sqrt{\frac{T}{μ}}\)
When length is doubled, then
v2=\(\frac{n_2}{2 \times2L}\sqrt{\frac{T}{μ}}\)
Dividing Eq.(i) by Eq. (ii), we get
\(\frac{v_1}{v_2}=\frac{n_1}{n_2}×2\)
To keep the resonance,
\(\frac{v_1}{v_2}=1=\frac{n_1}{n_2}×2\)
⇒n2=2n1
Hence, when the wire is doubled, the number of loops also get doubled to produce the resonance. That is, it resonates in second harmonic.
