Physics MCQs for NEET — Practice Questions with Answers

Time period in vibration magnetometer

             T=2$\mathrm{\pi }$$\sqrt{\frac{I}{M\times B_H}}$

$\Rightarrow$       T$\propto$$\frac{1}{\sqrt{B_H}}$

$\Rightarrow$       $\frac{T_1}{T_{2_{}}}$=$\sqrt{\frac{(B_H{)}_2}{(B_H{)}_1}}$

$\Rightarrow$       $\frac{2}{T_2}$=$\sqrt{\frac{6}{24}}$

             T = 4s

Electromagnetics is made of soft iron because soft iron has low retentivity and low coercive force or low coercivity. Soft iron is a soft magnetic material.

Atoms which have paired electrons have the magnetic moment zero. Because diamagnetism is the intrinsic property of every material and it is generated due to mutual interaction between the applied magnetic field orbital motion of electrons.

So, the magnetic moment of a diamagnetic atom is equal to zero.

Diamagnetic substance are weekly magnetised in a direction opposite to that of applied magnetic field. These are repelled in an external magnetic field i.e., have a tendency to move from high to low field region. 

The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy.

When magnetic dipole is rotated from initial position $\theta ={\theta }_1$ to final position $\theta ={\theta }_2,$ then work done =MB$\left(\cos {\theta }_1-\cos {\theta }_2\right)$

=MB$\left(1-\frac{1}{2}\right)$

=$\frac{2\times {10}^4\times 6\times {10}^{-4}}{2}=6J$

Ferromagnetism decreases with rise in temperature. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the curie temperature of the substance. 
Note: The Curie temperature of Iron is ${770}^{°}C$ and that of Nickel in ${358}^{°}C$.

3.

$\begin{array}{rl}& \mathrm{e}=-\frac{\mathrm{LdI}}{\mathrm{dt}}\\ & 30=\mathrm{L}\times \frac{6}{0.3}\\ & \mathrm{L}=1.5\mathrm{H}\end{array}$

$\mathrm{U}=\frac{1}{2}{\mathrm{LI}}^2$

$$\begin{gathered} \left(1\right) \\ \frac{{\mathrm{L}}_1}{{\mathrm{L}}_2} = \frac{{\mathrm{N}}_1^2}{{\mathrm{N}}_2^2} \\ \frac{36}{{\mathrm{L}}_2} = {\left(\frac{600}{500}\right)}^2 \\ {\mathrm{L}}_2 = 25 \mathrm{mH} \end{gathered}$$

$$\begin{gathered} \mathrm{e}=-\mathrm{L}\frac{\mathrm{dl}}{\mathrm{dt}} \\ =5\times 3=15 \mathrm{volt} \left(current is decreasding\right) \end{gathered}$$