CBSE 11th Standard Physics Subject Ncert Exemplar 1 Mark Questions 2021 Part - I
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CBSE 11th Standard Physics Subject Ncert Exemplar 1 Mark Questions 2021 Part - I
11th Standard CBSE
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Reg.No. :
Physics
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Suppose we use a physical balance to measure the mass of an object and find the mean value of our observation to be 156.28 g. Represent this result correctly.
(a)It least count of physical balance is 0.1 g, the mass measured will be correctly represented as
\(m=\left( 156.3\pm 0.1 \right) g\) -
(n-1) equal point masses each of mass m are palced at the verticle of a regular n-polygon. The vacant vartex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.
(a)Suppose, b be the position vector of centre of mass of regular n-polygon.
As (n-1) equal point masses each of mass m are placed at (n-1) vertices of regular polygon, therefore
\(\frac { (n-1)mb+ma }{ (n-1+1)m } =0\)
\(\\ \Rightarrow (n-1)mb+ma=0\)
\(\\ \Rightarrow b=\frac { -a }{ (n-1) } \) -
Out of aphelion and perihelion, where is the speed of the earth more and why?
(a)At perihelion, because the earth has to cover greater linear distance to keep the areal velocity constant.
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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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The volume of a given mass of a gas at 270C, 1 atm is 100 cc. What will be its volume at 3270C?
(a)Keeping p constant, we have
\({ V }_{ 2 }\) = \(\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } }\)
= \(\frac { 100\times 600 }{ 300 } \)
= 200cc -
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,
\(\lambda \propto \frac { 1 }{ { d }^{ 2 } } \)
\(Given,\ { d }_{ 1 }=1\overset { 0 }{ A } \ and\ { d }_{ 2 }=2\overset { 0 }{ A } \Rightarrow { \lambda }_{ 1 }:{ \lambda }_{ 2 }=4:1\) -
Under what condition is the motion of a simple pendulum be simple harmonic?
(a)When the displacement amplitude of the pendulum is extremely small as compared to its length.
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What is the ratio between the distance travelled by the oscillator iin one time period and amplitude?
(a)Total distance travelled by an oscillator in one time period, from its mean position to one extreme position, then to other extreme position and finally back to mean position is 4 A, where A is the amplitude of oscillation.
Hence, the ratio \(=\frac { 4A }{ A } =4\) -
What is the nature of water waves produced by a motorboat sailing in water ?
(a)Water waves produced by a motorboat sailing in water are both longitudinal and transverse.
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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 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 2021 Part - I Answer Keys
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It least count of physical balance is 0.1 g, the mass measured will be correctly represented as m=(156.3±0.1)g
It least count of physical balance is 0.1 g, the mass measured will be correctly represented as m=(156.3±0.1)g
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Suppose, b be the position vector of centre of mass of regular n-polygon.
As (n-1) equal point masses each of mass m are placed at (n-1) vertices of regular polygon, therefore
\( \frac{(n-1) m b+m a}{(n-1+1) m}=0 \)
\( \Rightarrow(n-1) m b+m a=0 \)
\( \Rightarrow b=\frac{-a}{(n-1)} \)Suppose, b be the position vector of centre of mass of regular n-polygon.
As (n-1) equal point masses each of mass m are placed at (n-1) vertices of regular polygon, therefore
\( \frac{(n-1) m b+m a}{(n-1+1) m}=0 \)
\( \Rightarrow(n-1) m b+m a=0 \)
\( \Rightarrow b=\frac{-a}{(n-1)} \) -
At perihelion, because the earth has to cover greater linear distance to keep the areal velocity constant.
At perihelion, because the earth has to cover greater linear distance to keep the areal velocity constant.
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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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Keeping p constant, we have
V2/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200ccKeeping p constant, we have
V2/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200cc -
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 -
When the displacement amplitude of the pendulum is extremely small as compared to its length.
When the displacement amplitude of the pendulum is extremely small as compared to its length.
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Total distance travelled by an oscillator in one time period, from its mean position to one extreme position, then to other extreme position and finally back to mean position is 4 A, where A is the amplitude of oscillation.
Hence, the ratio =4A/A=4Total distance travelled by an oscillator in one time period, from its mean position to one extreme position, then to other extreme position and finally back to mean position is 4 A, where A is the amplitude of oscillation.
Hence, the ratio =4A/A=4 -
Water waves produced by a motorboat sailing in water are both longitudinal and transverse.
Water waves produced by a motorboat sailing in water are both longitudinal and transverse.
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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 -
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.
