263846
A 5.5 m long string has a mass of 0.035 kg . If the tension in the string is 77 N , then the speed of a transverse wave on the string is :
1 \(770 \mathrm{~m} / \mathrm{sec}\).
2 \(120 \mathrm{~m} / \mathrm{sec}\).
3 \(175 \mathrm{~m} / \mathrm{sec}\).
4 \(110 \mathrm{~m} / \mathrm{sec}\).
Explanation:
d mass per unit length \(\mathrm{m}=\frac{0.035}{5.5}\) \[ =\frac{35}{5500}=\frac{7}{1100} \] so speed of transverse wave in string \(v=\sqrt{\frac{T}{m}}\) \[ \begin{aligned} & v=\sqrt{\frac{77}{7} \times 1100}=\sqrt{(11)^2 \times 100} \\ & v=110 \mathrm{~m} / \mathrm{sec} \end{aligned} \]
NCERT-XI-II-285
TEST SERIES (PHYSICS FST)
263835
In an electromagnetic wave in free space the root mean square value of the electric field is \(\mathrm{E}_{\mathrm{mm}}=6 \mathrm{~V} / \mathrm{m}\). The peak value of the magnetic field is:
263836
An electric dipole of moment \(p\) is placed in the position of stable equilibrium in uniform electric field of intensity E. This is rotated through an angle \(\theta\) from the initial position. The potential energy of the electric dipole in the final position is :
1 \(-\mathrm{pE} \cos \theta\)
2 \(\mathrm{pE}(1-\cos \theta)\)
3 \(\mathrm{pE} \cos \theta\)
4 \(\mathrm{pE} \sin \theta\).
Explanation:
a \(\therefore\) Potential Energy \(=-p \mathrm{E} \cos \theta\)
NCERT-XII-I-60
TEST SERIES (PHYSICS FST)
263837
A proton carrying 1 HeV kinetic energy is moving in a circular path of radius \(R\) in uniform magnetic field. What should be the energy of an a-particle to describe a circle of same radius in the same field:
1 2 MeV
2 1 MeV
3 0.5 MeV
4 4 MeV .
Explanation:
b Knetic energy of a charged particle. \[ \begin{aligned} & K=\frac{1}{2} m v^2 \text { or } v=\sqrt{\frac{2 K}{m}} \\ & R=\frac{m v}{B q}=\frac{m}{B q} \sqrt{\frac{2 K}{m}}=\frac{\sqrt{2 m K}}{B q} \\ & \therefore R_p=\frac{\sqrt{2 m_p K_p}}{B q_p}=\frac{\sqrt{2 m K_p}}{B e} \\ & \text { and } R_a=\frac{\sqrt{2 m_p K_a}}{B q_a}=\frac{\sqrt{2(4 m) K_a}}{B(2 e)}=\frac{\sqrt{2 m K_a}}{B e} \\ & \therefore \frac{R_p}{R_a}=\sqrt{\frac{K_p}{K_a}} \end{aligned} \] \(\mathrm{As}, \mathrm{R}_{\mathrm{p}}=\mathrm{R}_{\mathrm{a}}\) (given) \[ \therefore K_{\mathrm{a}}=\mathrm{K}_{\mathrm{p}}=1 \mathrm{MeV} . \]
NEET Test Series from KOTA - 10 Papers In MS WORD
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TEST SERIES (PHYSICS FST)
263846
A 5.5 m long string has a mass of 0.035 kg . If the tension in the string is 77 N , then the speed of a transverse wave on the string is :
1 \(770 \mathrm{~m} / \mathrm{sec}\).
2 \(120 \mathrm{~m} / \mathrm{sec}\).
3 \(175 \mathrm{~m} / \mathrm{sec}\).
4 \(110 \mathrm{~m} / \mathrm{sec}\).
Explanation:
d mass per unit length \(\mathrm{m}=\frac{0.035}{5.5}\) \[ =\frac{35}{5500}=\frac{7}{1100} \] so speed of transverse wave in string \(v=\sqrt{\frac{T}{m}}\) \[ \begin{aligned} & v=\sqrt{\frac{77}{7} \times 1100}=\sqrt{(11)^2 \times 100} \\ & v=110 \mathrm{~m} / \mathrm{sec} \end{aligned} \]
NCERT-XI-II-285
TEST SERIES (PHYSICS FST)
263835
In an electromagnetic wave in free space the root mean square value of the electric field is \(\mathrm{E}_{\mathrm{mm}}=6 \mathrm{~V} / \mathrm{m}\). The peak value of the magnetic field is:
263836
An electric dipole of moment \(p\) is placed in the position of stable equilibrium in uniform electric field of intensity E. This is rotated through an angle \(\theta\) from the initial position. The potential energy of the electric dipole in the final position is :
1 \(-\mathrm{pE} \cos \theta\)
2 \(\mathrm{pE}(1-\cos \theta)\)
3 \(\mathrm{pE} \cos \theta\)
4 \(\mathrm{pE} \sin \theta\).
Explanation:
a \(\therefore\) Potential Energy \(=-p \mathrm{E} \cos \theta\)
NCERT-XII-I-60
TEST SERIES (PHYSICS FST)
263837
A proton carrying 1 HeV kinetic energy is moving in a circular path of radius \(R\) in uniform magnetic field. What should be the energy of an a-particle to describe a circle of same radius in the same field:
1 2 MeV
2 1 MeV
3 0.5 MeV
4 4 MeV .
Explanation:
b Knetic energy of a charged particle. \[ \begin{aligned} & K=\frac{1}{2} m v^2 \text { or } v=\sqrt{\frac{2 K}{m}} \\ & R=\frac{m v}{B q}=\frac{m}{B q} \sqrt{\frac{2 K}{m}}=\frac{\sqrt{2 m K}}{B q} \\ & \therefore R_p=\frac{\sqrt{2 m_p K_p}}{B q_p}=\frac{\sqrt{2 m K_p}}{B e} \\ & \text { and } R_a=\frac{\sqrt{2 m_p K_a}}{B q_a}=\frac{\sqrt{2(4 m) K_a}}{B(2 e)}=\frac{\sqrt{2 m K_a}}{B e} \\ & \therefore \frac{R_p}{R_a}=\sqrt{\frac{K_p}{K_a}} \end{aligned} \] \(\mathrm{As}, \mathrm{R}_{\mathrm{p}}=\mathrm{R}_{\mathrm{a}}\) (given) \[ \therefore K_{\mathrm{a}}=\mathrm{K}_{\mathrm{p}}=1 \mathrm{MeV} . \]
263846
A 5.5 m long string has a mass of 0.035 kg . If the tension in the string is 77 N , then the speed of a transverse wave on the string is :
1 \(770 \mathrm{~m} / \mathrm{sec}\).
2 \(120 \mathrm{~m} / \mathrm{sec}\).
3 \(175 \mathrm{~m} / \mathrm{sec}\).
4 \(110 \mathrm{~m} / \mathrm{sec}\).
Explanation:
d mass per unit length \(\mathrm{m}=\frac{0.035}{5.5}\) \[ =\frac{35}{5500}=\frac{7}{1100} \] so speed of transverse wave in string \(v=\sqrt{\frac{T}{m}}\) \[ \begin{aligned} & v=\sqrt{\frac{77}{7} \times 1100}=\sqrt{(11)^2 \times 100} \\ & v=110 \mathrm{~m} / \mathrm{sec} \end{aligned} \]
NCERT-XI-II-285
TEST SERIES (PHYSICS FST)
263835
In an electromagnetic wave in free space the root mean square value of the electric field is \(\mathrm{E}_{\mathrm{mm}}=6 \mathrm{~V} / \mathrm{m}\). The peak value of the magnetic field is:
263836
An electric dipole of moment \(p\) is placed in the position of stable equilibrium in uniform electric field of intensity E. This is rotated through an angle \(\theta\) from the initial position. The potential energy of the electric dipole in the final position is :
1 \(-\mathrm{pE} \cos \theta\)
2 \(\mathrm{pE}(1-\cos \theta)\)
3 \(\mathrm{pE} \cos \theta\)
4 \(\mathrm{pE} \sin \theta\).
Explanation:
a \(\therefore\) Potential Energy \(=-p \mathrm{E} \cos \theta\)
NCERT-XII-I-60
TEST SERIES (PHYSICS FST)
263837
A proton carrying 1 HeV kinetic energy is moving in a circular path of radius \(R\) in uniform magnetic field. What should be the energy of an a-particle to describe a circle of same radius in the same field:
1 2 MeV
2 1 MeV
3 0.5 MeV
4 4 MeV .
Explanation:
b Knetic energy of a charged particle. \[ \begin{aligned} & K=\frac{1}{2} m v^2 \text { or } v=\sqrt{\frac{2 K}{m}} \\ & R=\frac{m v}{B q}=\frac{m}{B q} \sqrt{\frac{2 K}{m}}=\frac{\sqrt{2 m K}}{B q} \\ & \therefore R_p=\frac{\sqrt{2 m_p K_p}}{B q_p}=\frac{\sqrt{2 m K_p}}{B e} \\ & \text { and } R_a=\frac{\sqrt{2 m_p K_a}}{B q_a}=\frac{\sqrt{2(4 m) K_a}}{B(2 e)}=\frac{\sqrt{2 m K_a}}{B e} \\ & \therefore \frac{R_p}{R_a}=\sqrt{\frac{K_p}{K_a}} \end{aligned} \] \(\mathrm{As}, \mathrm{R}_{\mathrm{p}}=\mathrm{R}_{\mathrm{a}}\) (given) \[ \therefore K_{\mathrm{a}}=\mathrm{K}_{\mathrm{p}}=1 \mathrm{MeV} . \]
263846
A 5.5 m long string has a mass of 0.035 kg . If the tension in the string is 77 N , then the speed of a transverse wave on the string is :
1 \(770 \mathrm{~m} / \mathrm{sec}\).
2 \(120 \mathrm{~m} / \mathrm{sec}\).
3 \(175 \mathrm{~m} / \mathrm{sec}\).
4 \(110 \mathrm{~m} / \mathrm{sec}\).
Explanation:
d mass per unit length \(\mathrm{m}=\frac{0.035}{5.5}\) \[ =\frac{35}{5500}=\frac{7}{1100} \] so speed of transverse wave in string \(v=\sqrt{\frac{T}{m}}\) \[ \begin{aligned} & v=\sqrt{\frac{77}{7} \times 1100}=\sqrt{(11)^2 \times 100} \\ & v=110 \mathrm{~m} / \mathrm{sec} \end{aligned} \]
NCERT-XI-II-285
TEST SERIES (PHYSICS FST)
263835
In an electromagnetic wave in free space the root mean square value of the electric field is \(\mathrm{E}_{\mathrm{mm}}=6 \mathrm{~V} / \mathrm{m}\). The peak value of the magnetic field is:
263836
An electric dipole of moment \(p\) is placed in the position of stable equilibrium in uniform electric field of intensity E. This is rotated through an angle \(\theta\) from the initial position. The potential energy of the electric dipole in the final position is :
1 \(-\mathrm{pE} \cos \theta\)
2 \(\mathrm{pE}(1-\cos \theta)\)
3 \(\mathrm{pE} \cos \theta\)
4 \(\mathrm{pE} \sin \theta\).
Explanation:
a \(\therefore\) Potential Energy \(=-p \mathrm{E} \cos \theta\)
NCERT-XII-I-60
TEST SERIES (PHYSICS FST)
263837
A proton carrying 1 HeV kinetic energy is moving in a circular path of radius \(R\) in uniform magnetic field. What should be the energy of an a-particle to describe a circle of same radius in the same field:
1 2 MeV
2 1 MeV
3 0.5 MeV
4 4 MeV .
Explanation:
b Knetic energy of a charged particle. \[ \begin{aligned} & K=\frac{1}{2} m v^2 \text { or } v=\sqrt{\frac{2 K}{m}} \\ & R=\frac{m v}{B q}=\frac{m}{B q} \sqrt{\frac{2 K}{m}}=\frac{\sqrt{2 m K}}{B q} \\ & \therefore R_p=\frac{\sqrt{2 m_p K_p}}{B q_p}=\frac{\sqrt{2 m K_p}}{B e} \\ & \text { and } R_a=\frac{\sqrt{2 m_p K_a}}{B q_a}=\frac{\sqrt{2(4 m) K_a}}{B(2 e)}=\frac{\sqrt{2 m K_a}}{B e} \\ & \therefore \frac{R_p}{R_a}=\sqrt{\frac{K_p}{K_a}} \end{aligned} \] \(\mathrm{As}, \mathrm{R}_{\mathrm{p}}=\mathrm{R}_{\mathrm{a}}\) (given) \[ \therefore K_{\mathrm{a}}=\mathrm{K}_{\mathrm{p}}=1 \mathrm{MeV} . \]
