266501
A body travels uniformly a distance of \((13.8 \pm 0.2) \mathrm{m}\) in a time \((4.0 \pm 0.3) \mathrm{s}\). The velocity of the body within error limits is
1 \((3.45 \pm 0.2) \mathrm{ms}^{-1}\)
2 \((3.45 \pm 0.3) \mathrm{ms}^{-1}\)
3 \((3.45 \pm 0.4) \mathrm{ms}^{-1}\)
4 \((3.45 \pm 0.5) \mathrm{ms}^{-1}\)
Explanation:
b Percentage error, we have \[ \begin{aligned} S & =13.8 \pm \frac{0.2}{13.8} \times 100 \\ & =13.8 \pm 1.4 \% \\ t & =4.0 \pm \frac{0.3}{4} \times 100 \\ & =4 \pm 7.5 \% \\ v & =\frac{s}{t}=\frac{13.8 \pm 1.4}{4 \pm 7.5} \quad=(3.45 \pm 0.3) \mathrm{ms}^{-1} \end{aligned} \]
**NHC**
TEST SERIES (PHYSICS FST)
266502
Two tuning fouks of frequencies 256 and 258 vibrations/sec are sounded together, the time interval between consecutive maxima heard by the observer is
1 2 sec
2 0.5 sec
3 250 sec
4 252 sec
Explanation:
b \[ t=\frac{1}{258-256}=\frac{1}{2}=0.5 \mathrm{sec} . \]
**NCERT-XI-II-293**
TEST SERIES (PHYSICS FST)
266503
The ratio of the lengths of two wires \(A\) and \(B\) of same material is \(1: 2\) and the ratio of their diameter is \(2: 1\). They are stretched by the same force, then the ratio of increase in length will be
266504
An arc of a circle of radius \(R\) subtends an angle \(r / 2\) at the centre. It carries a current \(\mathbf{i}\). The magnetic field at the centre will be
1 \(\frac{\mu_0 i}{2 R}\)
2 \(\frac{\mu_0 \mathrm{i}}{8 R}\)
3 \(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\)
4 \(\frac{2 \mu_0 i}{5 R}\)
Explanation:
b
TEST SERIES (PHYSICS FST)
266505
If \(\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3\) are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 \(E_1>E_3>E_2\)
2 \(E_2>E_3>E_1\)
3 \(E_1>E_2>E_3\)
4 \(E_1=E_2=E_3\)
Explanation:
a According to relation \[ \begin{aligned} & E=\frac{1}{2} m v^2 \\ & v=\sqrt{\frac{2 E}{m}} \\ & \lambda=\frac{h}{\sqrt{2 m E}} \end{aligned} \] Because \(\mathrm{m}_1<\mathrm{m}_3<\mathrm{m}_2\) So for same \(\lambda_1 E_1>E_3>E_2\)
266501
A body travels uniformly a distance of \((13.8 \pm 0.2) \mathrm{m}\) in a time \((4.0 \pm 0.3) \mathrm{s}\). The velocity of the body within error limits is
1 \((3.45 \pm 0.2) \mathrm{ms}^{-1}\)
2 \((3.45 \pm 0.3) \mathrm{ms}^{-1}\)
3 \((3.45 \pm 0.4) \mathrm{ms}^{-1}\)
4 \((3.45 \pm 0.5) \mathrm{ms}^{-1}\)
Explanation:
b Percentage error, we have \[ \begin{aligned} S & =13.8 \pm \frac{0.2}{13.8} \times 100 \\ & =13.8 \pm 1.4 \% \\ t & =4.0 \pm \frac{0.3}{4} \times 100 \\ & =4 \pm 7.5 \% \\ v & =\frac{s}{t}=\frac{13.8 \pm 1.4}{4 \pm 7.5} \quad=(3.45 \pm 0.3) \mathrm{ms}^{-1} \end{aligned} \]
**NHC**
TEST SERIES (PHYSICS FST)
266502
Two tuning fouks of frequencies 256 and 258 vibrations/sec are sounded together, the time interval between consecutive maxima heard by the observer is
1 2 sec
2 0.5 sec
3 250 sec
4 252 sec
Explanation:
b \[ t=\frac{1}{258-256}=\frac{1}{2}=0.5 \mathrm{sec} . \]
**NCERT-XI-II-293**
TEST SERIES (PHYSICS FST)
266503
The ratio of the lengths of two wires \(A\) and \(B\) of same material is \(1: 2\) and the ratio of their diameter is \(2: 1\). They are stretched by the same force, then the ratio of increase in length will be
266504
An arc of a circle of radius \(R\) subtends an angle \(r / 2\) at the centre. It carries a current \(\mathbf{i}\). The magnetic field at the centre will be
1 \(\frac{\mu_0 i}{2 R}\)
2 \(\frac{\mu_0 \mathrm{i}}{8 R}\)
3 \(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\)
4 \(\frac{2 \mu_0 i}{5 R}\)
Explanation:
b
TEST SERIES (PHYSICS FST)
266505
If \(\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3\) are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 \(E_1>E_3>E_2\)
2 \(E_2>E_3>E_1\)
3 \(E_1>E_2>E_3\)
4 \(E_1=E_2=E_3\)
Explanation:
a According to relation \[ \begin{aligned} & E=\frac{1}{2} m v^2 \\ & v=\sqrt{\frac{2 E}{m}} \\ & \lambda=\frac{h}{\sqrt{2 m E}} \end{aligned} \] Because \(\mathrm{m}_1<\mathrm{m}_3<\mathrm{m}_2\) So for same \(\lambda_1 E_1>E_3>E_2\)
266501
A body travels uniformly a distance of \((13.8 \pm 0.2) \mathrm{m}\) in a time \((4.0 \pm 0.3) \mathrm{s}\). The velocity of the body within error limits is
1 \((3.45 \pm 0.2) \mathrm{ms}^{-1}\)
2 \((3.45 \pm 0.3) \mathrm{ms}^{-1}\)
3 \((3.45 \pm 0.4) \mathrm{ms}^{-1}\)
4 \((3.45 \pm 0.5) \mathrm{ms}^{-1}\)
Explanation:
b Percentage error, we have \[ \begin{aligned} S & =13.8 \pm \frac{0.2}{13.8} \times 100 \\ & =13.8 \pm 1.4 \% \\ t & =4.0 \pm \frac{0.3}{4} \times 100 \\ & =4 \pm 7.5 \% \\ v & =\frac{s}{t}=\frac{13.8 \pm 1.4}{4 \pm 7.5} \quad=(3.45 \pm 0.3) \mathrm{ms}^{-1} \end{aligned} \]
**NHC**
TEST SERIES (PHYSICS FST)
266502
Two tuning fouks of frequencies 256 and 258 vibrations/sec are sounded together, the time interval between consecutive maxima heard by the observer is
1 2 sec
2 0.5 sec
3 250 sec
4 252 sec
Explanation:
b \[ t=\frac{1}{258-256}=\frac{1}{2}=0.5 \mathrm{sec} . \]
**NCERT-XI-II-293**
TEST SERIES (PHYSICS FST)
266503
The ratio of the lengths of two wires \(A\) and \(B\) of same material is \(1: 2\) and the ratio of their diameter is \(2: 1\). They are stretched by the same force, then the ratio of increase in length will be
266504
An arc of a circle of radius \(R\) subtends an angle \(r / 2\) at the centre. It carries a current \(\mathbf{i}\). The magnetic field at the centre will be
1 \(\frac{\mu_0 i}{2 R}\)
2 \(\frac{\mu_0 \mathrm{i}}{8 R}\)
3 \(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\)
4 \(\frac{2 \mu_0 i}{5 R}\)
Explanation:
b
TEST SERIES (PHYSICS FST)
266505
If \(\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3\) are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 \(E_1>E_3>E_2\)
2 \(E_2>E_3>E_1\)
3 \(E_1>E_2>E_3\)
4 \(E_1=E_2=E_3\)
Explanation:
a According to relation \[ \begin{aligned} & E=\frac{1}{2} m v^2 \\ & v=\sqrt{\frac{2 E}{m}} \\ & \lambda=\frac{h}{\sqrt{2 m E}} \end{aligned} \] Because \(\mathrm{m}_1<\mathrm{m}_3<\mathrm{m}_2\) So for same \(\lambda_1 E_1>E_3>E_2\)
266501
A body travels uniformly a distance of \((13.8 \pm 0.2) \mathrm{m}\) in a time \((4.0 \pm 0.3) \mathrm{s}\). The velocity of the body within error limits is
1 \((3.45 \pm 0.2) \mathrm{ms}^{-1}\)
2 \((3.45 \pm 0.3) \mathrm{ms}^{-1}\)
3 \((3.45 \pm 0.4) \mathrm{ms}^{-1}\)
4 \((3.45 \pm 0.5) \mathrm{ms}^{-1}\)
Explanation:
b Percentage error, we have \[ \begin{aligned} S & =13.8 \pm \frac{0.2}{13.8} \times 100 \\ & =13.8 \pm 1.4 \% \\ t & =4.0 \pm \frac{0.3}{4} \times 100 \\ & =4 \pm 7.5 \% \\ v & =\frac{s}{t}=\frac{13.8 \pm 1.4}{4 \pm 7.5} \quad=(3.45 \pm 0.3) \mathrm{ms}^{-1} \end{aligned} \]
**NHC**
TEST SERIES (PHYSICS FST)
266502
Two tuning fouks of frequencies 256 and 258 vibrations/sec are sounded together, the time interval between consecutive maxima heard by the observer is
1 2 sec
2 0.5 sec
3 250 sec
4 252 sec
Explanation:
b \[ t=\frac{1}{258-256}=\frac{1}{2}=0.5 \mathrm{sec} . \]
**NCERT-XI-II-293**
TEST SERIES (PHYSICS FST)
266503
The ratio of the lengths of two wires \(A\) and \(B\) of same material is \(1: 2\) and the ratio of their diameter is \(2: 1\). They are stretched by the same force, then the ratio of increase in length will be
266504
An arc of a circle of radius \(R\) subtends an angle \(r / 2\) at the centre. It carries a current \(\mathbf{i}\). The magnetic field at the centre will be
1 \(\frac{\mu_0 i}{2 R}\)
2 \(\frac{\mu_0 \mathrm{i}}{8 R}\)
3 \(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\)
4 \(\frac{2 \mu_0 i}{5 R}\)
Explanation:
b
TEST SERIES (PHYSICS FST)
266505
If \(\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3\) are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 \(E_1>E_3>E_2\)
2 \(E_2>E_3>E_1\)
3 \(E_1>E_2>E_3\)
4 \(E_1=E_2=E_3\)
Explanation:
a According to relation \[ \begin{aligned} & E=\frac{1}{2} m v^2 \\ & v=\sqrt{\frac{2 E}{m}} \\ & \lambda=\frac{h}{\sqrt{2 m E}} \end{aligned} \] Because \(\mathrm{m}_1<\mathrm{m}_3<\mathrm{m}_2\) So for same \(\lambda_1 E_1>E_3>E_2\)
266501
A body travels uniformly a distance of \((13.8 \pm 0.2) \mathrm{m}\) in a time \((4.0 \pm 0.3) \mathrm{s}\). The velocity of the body within error limits is
1 \((3.45 \pm 0.2) \mathrm{ms}^{-1}\)
2 \((3.45 \pm 0.3) \mathrm{ms}^{-1}\)
3 \((3.45 \pm 0.4) \mathrm{ms}^{-1}\)
4 \((3.45 \pm 0.5) \mathrm{ms}^{-1}\)
Explanation:
b Percentage error, we have \[ \begin{aligned} S & =13.8 \pm \frac{0.2}{13.8} \times 100 \\ & =13.8 \pm 1.4 \% \\ t & =4.0 \pm \frac{0.3}{4} \times 100 \\ & =4 \pm 7.5 \% \\ v & =\frac{s}{t}=\frac{13.8 \pm 1.4}{4 \pm 7.5} \quad=(3.45 \pm 0.3) \mathrm{ms}^{-1} \end{aligned} \]
**NHC**
TEST SERIES (PHYSICS FST)
266502
Two tuning fouks of frequencies 256 and 258 vibrations/sec are sounded together, the time interval between consecutive maxima heard by the observer is
1 2 sec
2 0.5 sec
3 250 sec
4 252 sec
Explanation:
b \[ t=\frac{1}{258-256}=\frac{1}{2}=0.5 \mathrm{sec} . \]
**NCERT-XI-II-293**
TEST SERIES (PHYSICS FST)
266503
The ratio of the lengths of two wires \(A\) and \(B\) of same material is \(1: 2\) and the ratio of their diameter is \(2: 1\). They are stretched by the same force, then the ratio of increase in length will be
266504
An arc of a circle of radius \(R\) subtends an angle \(r / 2\) at the centre. It carries a current \(\mathbf{i}\). The magnetic field at the centre will be
1 \(\frac{\mu_0 i}{2 R}\)
2 \(\frac{\mu_0 \mathrm{i}}{8 R}\)
3 \(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\)
4 \(\frac{2 \mu_0 i}{5 R}\)
Explanation:
b
TEST SERIES (PHYSICS FST)
266505
If \(\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3\) are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 \(E_1>E_3>E_2\)
2 \(E_2>E_3>E_1\)
3 \(E_1>E_2>E_3\)
4 \(E_1=E_2=E_3\)
Explanation:
a According to relation \[ \begin{aligned} & E=\frac{1}{2} m v^2 \\ & v=\sqrt{\frac{2 E}{m}} \\ & \lambda=\frac{h}{\sqrt{2 m E}} \end{aligned} \] Because \(\mathrm{m}_1<\mathrm{m}_3<\mathrm{m}_2\) So for same \(\lambda_1 E_1>E_3>E_2\)
