263810
Ratio of magnetic fields at the centre of a current carrying coil of radius \(R\) and at distance of \(3 R\) on its axis is :
1 \(10 \sqrt{10}\)
2 \(20 \sqrt{10}\)
3 \(2 \sqrt{10}\)
4 \(\sqrt{10}\)
Explanation:
a The desired ratio is \[ \frac{\frac{\mu_0 i}{2 R}}{\frac{\mu_{\mathrm{c}} \mathrm{R}^2}{2\left(R^2+9 R^2\right)^{3 / 2}}}=10 \sqrt{10} \]
NCERT-XII-I-115
TEST SERIES (PHYSICS FST)
263811
A dia-magnetic material in a magnetic field moves:
1 Perpendicular to the field
2 From weakerto stronger parts
3 From stronger to weaker parts
4 In пone of the above directions
Explanation:
c We know that, diamagnetic materials are magnetised in the opposite direction of the magnetising field. Therefore, diamagnetic matrerial moves from stronger to weaker part of the magnetic fields.
NCERT-XII-I-197
TEST SERIES (PHYSICS FST)
263812
A stone is thrown at \(25 \mathrm{~m} / \mathrm{s}\) at \(53^{\circ}\) above the horizontal. At what time its velocity is at \(45^{\circ}\) below the horizontal :
1 0.5 s
2 4 s
3 3.5 s
4 2.5 s
Explanation:
c Horizontal component of velcoity, throughout the motion remain constant \[ \begin{aligned} & u \cos \theta=v \cos \alpha \\ & 25 \cos 53^{\circ}=v \cos 45^{\circ} \\ & \Rightarrow v=25 \times \frac{3}{5} \cdot \sqrt{2}=15 \sqrt{2} \\ & \Rightarrow v_y=v \sin 45=15 \end{aligned} \] Now, using \(V_y=u_y+a_y t\) in \(y\)-direction \[ \begin{aligned} & -15=25 \sin 53^{\circ}-g t \\ & -15=25 \times \frac{4}{5}-10 \mathrm{t} \\ & t=3.5 \mathrm{sec} . \end{aligned} \]
NCERT-XII-I-39
TEST SERIES (PHYSICS FST)
263813
The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :
263810
Ratio of magnetic fields at the centre of a current carrying coil of radius \(R\) and at distance of \(3 R\) on its axis is :
1 \(10 \sqrt{10}\)
2 \(20 \sqrt{10}\)
3 \(2 \sqrt{10}\)
4 \(\sqrt{10}\)
Explanation:
a The desired ratio is \[ \frac{\frac{\mu_0 i}{2 R}}{\frac{\mu_{\mathrm{c}} \mathrm{R}^2}{2\left(R^2+9 R^2\right)^{3 / 2}}}=10 \sqrt{10} \]
NCERT-XII-I-115
TEST SERIES (PHYSICS FST)
263811
A dia-magnetic material in a magnetic field moves:
1 Perpendicular to the field
2 From weakerto stronger parts
3 From stronger to weaker parts
4 In пone of the above directions
Explanation:
c We know that, diamagnetic materials are magnetised in the opposite direction of the magnetising field. Therefore, diamagnetic matrerial moves from stronger to weaker part of the magnetic fields.
NCERT-XII-I-197
TEST SERIES (PHYSICS FST)
263812
A stone is thrown at \(25 \mathrm{~m} / \mathrm{s}\) at \(53^{\circ}\) above the horizontal. At what time its velocity is at \(45^{\circ}\) below the horizontal :
1 0.5 s
2 4 s
3 3.5 s
4 2.5 s
Explanation:
c Horizontal component of velcoity, throughout the motion remain constant \[ \begin{aligned} & u \cos \theta=v \cos \alpha \\ & 25 \cos 53^{\circ}=v \cos 45^{\circ} \\ & \Rightarrow v=25 \times \frac{3}{5} \cdot \sqrt{2}=15 \sqrt{2} \\ & \Rightarrow v_y=v \sin 45=15 \end{aligned} \] Now, using \(V_y=u_y+a_y t\) in \(y\)-direction \[ \begin{aligned} & -15=25 \sin 53^{\circ}-g t \\ & -15=25 \times \frac{4}{5}-10 \mathrm{t} \\ & t=3.5 \mathrm{sec} . \end{aligned} \]
NCERT-XII-I-39
TEST SERIES (PHYSICS FST)
263813
The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :
263810
Ratio of magnetic fields at the centre of a current carrying coil of radius \(R\) and at distance of \(3 R\) on its axis is :
1 \(10 \sqrt{10}\)
2 \(20 \sqrt{10}\)
3 \(2 \sqrt{10}\)
4 \(\sqrt{10}\)
Explanation:
a The desired ratio is \[ \frac{\frac{\mu_0 i}{2 R}}{\frac{\mu_{\mathrm{c}} \mathrm{R}^2}{2\left(R^2+9 R^2\right)^{3 / 2}}}=10 \sqrt{10} \]
NCERT-XII-I-115
TEST SERIES (PHYSICS FST)
263811
A dia-magnetic material in a magnetic field moves:
1 Perpendicular to the field
2 From weakerto stronger parts
3 From stronger to weaker parts
4 In пone of the above directions
Explanation:
c We know that, diamagnetic materials are magnetised in the opposite direction of the magnetising field. Therefore, diamagnetic matrerial moves from stronger to weaker part of the magnetic fields.
NCERT-XII-I-197
TEST SERIES (PHYSICS FST)
263812
A stone is thrown at \(25 \mathrm{~m} / \mathrm{s}\) at \(53^{\circ}\) above the horizontal. At what time its velocity is at \(45^{\circ}\) below the horizontal :
1 0.5 s
2 4 s
3 3.5 s
4 2.5 s
Explanation:
c Horizontal component of velcoity, throughout the motion remain constant \[ \begin{aligned} & u \cos \theta=v \cos \alpha \\ & 25 \cos 53^{\circ}=v \cos 45^{\circ} \\ & \Rightarrow v=25 \times \frac{3}{5} \cdot \sqrt{2}=15 \sqrt{2} \\ & \Rightarrow v_y=v \sin 45=15 \end{aligned} \] Now, using \(V_y=u_y+a_y t\) in \(y\)-direction \[ \begin{aligned} & -15=25 \sin 53^{\circ}-g t \\ & -15=25 \times \frac{4}{5}-10 \mathrm{t} \\ & t=3.5 \mathrm{sec} . \end{aligned} \]
NCERT-XII-I-39
TEST SERIES (PHYSICS FST)
263813
The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :
263810
Ratio of magnetic fields at the centre of a current carrying coil of radius \(R\) and at distance of \(3 R\) on its axis is :
1 \(10 \sqrt{10}\)
2 \(20 \sqrt{10}\)
3 \(2 \sqrt{10}\)
4 \(\sqrt{10}\)
Explanation:
a The desired ratio is \[ \frac{\frac{\mu_0 i}{2 R}}{\frac{\mu_{\mathrm{c}} \mathrm{R}^2}{2\left(R^2+9 R^2\right)^{3 / 2}}}=10 \sqrt{10} \]
NCERT-XII-I-115
TEST SERIES (PHYSICS FST)
263811
A dia-magnetic material in a magnetic field moves:
1 Perpendicular to the field
2 From weakerto stronger parts
3 From stronger to weaker parts
4 In пone of the above directions
Explanation:
c We know that, diamagnetic materials are magnetised in the opposite direction of the magnetising field. Therefore, diamagnetic matrerial moves from stronger to weaker part of the magnetic fields.
NCERT-XII-I-197
TEST SERIES (PHYSICS FST)
263812
A stone is thrown at \(25 \mathrm{~m} / \mathrm{s}\) at \(53^{\circ}\) above the horizontal. At what time its velocity is at \(45^{\circ}\) below the horizontal :
1 0.5 s
2 4 s
3 3.5 s
4 2.5 s
Explanation:
c Horizontal component of velcoity, throughout the motion remain constant \[ \begin{aligned} & u \cos \theta=v \cos \alpha \\ & 25 \cos 53^{\circ}=v \cos 45^{\circ} \\ & \Rightarrow v=25 \times \frac{3}{5} \cdot \sqrt{2}=15 \sqrt{2} \\ & \Rightarrow v_y=v \sin 45=15 \end{aligned} \] Now, using \(V_y=u_y+a_y t\) in \(y\)-direction \[ \begin{aligned} & -15=25 \sin 53^{\circ}-g t \\ & -15=25 \times \frac{4}{5}-10 \mathrm{t} \\ & t=3.5 \mathrm{sec} . \end{aligned} \]
NCERT-XII-I-39
TEST SERIES (PHYSICS FST)
263813
The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :
263810
Ratio of magnetic fields at the centre of a current carrying coil of radius \(R\) and at distance of \(3 R\) on its axis is :
1 \(10 \sqrt{10}\)
2 \(20 \sqrt{10}\)
3 \(2 \sqrt{10}\)
4 \(\sqrt{10}\)
Explanation:
a The desired ratio is \[ \frac{\frac{\mu_0 i}{2 R}}{\frac{\mu_{\mathrm{c}} \mathrm{R}^2}{2\left(R^2+9 R^2\right)^{3 / 2}}}=10 \sqrt{10} \]
NCERT-XII-I-115
TEST SERIES (PHYSICS FST)
263811
A dia-magnetic material in a magnetic field moves:
1 Perpendicular to the field
2 From weakerto stronger parts
3 From stronger to weaker parts
4 In пone of the above directions
Explanation:
c We know that, diamagnetic materials are magnetised in the opposite direction of the magnetising field. Therefore, diamagnetic matrerial moves from stronger to weaker part of the magnetic fields.
NCERT-XII-I-197
TEST SERIES (PHYSICS FST)
263812
A stone is thrown at \(25 \mathrm{~m} / \mathrm{s}\) at \(53^{\circ}\) above the horizontal. At what time its velocity is at \(45^{\circ}\) below the horizontal :
1 0.5 s
2 4 s
3 3.5 s
4 2.5 s
Explanation:
c Horizontal component of velcoity, throughout the motion remain constant \[ \begin{aligned} & u \cos \theta=v \cos \alpha \\ & 25 \cos 53^{\circ}=v \cos 45^{\circ} \\ & \Rightarrow v=25 \times \frac{3}{5} \cdot \sqrt{2}=15 \sqrt{2} \\ & \Rightarrow v_y=v \sin 45=15 \end{aligned} \] Now, using \(V_y=u_y+a_y t\) in \(y\)-direction \[ \begin{aligned} & -15=25 \sin 53^{\circ}-g t \\ & -15=25 \times \frac{4}{5}-10 \mathrm{t} \\ & t=3.5 \mathrm{sec} . \end{aligned} \]
NCERT-XII-I-39
TEST SERIES (PHYSICS FST)
263813
The ratio of the distance covered to the displacement of a particle moved along a semi-circle of radius ris :
