Question 14 Marks
Answer
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Sol.
$\begin{array}{l} R _{ eq }=2 \Omega \\ I =\frac{2}{2}=1 A\end{array}$
Sol.
$\begin{array}{l} R _{ eq }=2 \Omega \\ I =\frac{2}{2}=1 A\end{array}$
Questions
SECTION - B [PHYSICS - NUMERIC]
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5 questions · self-marked practice — reveal the answer and mark yourself.
Question 24 Marks
A tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is F then angular velocity of the tube is $\sqrt{\frac{F}{\alpha M}}$ in SI unit. The value of $\alpha$ is _________.
Answer
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Sol.
$\begin{array}{l} F =2 M \omega^2 \frac{\ell}{2}= Mw ^2 \ell \\ \omega=\sqrt{\frac{ F }{ M \ell}}\end{array}$
Sol.
$\begin{array}{l} F =2 M \omega^2 \frac{\ell}{2}= Mw ^2 \ell \\ \omega=\sqrt{\frac{ F }{ M \ell}}\end{array}$
Question 34 Marks
A parallel plate capacitor of area $A =16 cm^2$ and separation between the plates 10 cm, is charged by a DC current. Consider a hypothetical plane surface of area $A_0=3.2 cm^2$ inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6 A . At the same instant the displacement current through $A _0$ is _________ mA .
Answer
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Sol. $J_d=\frac{I}{A}=\frac{6}{16}$
$\therefore$ I through small area $=J_d \times A^{\prime}=\frac{6}{16} \times 3.2=$$
1.2 A=1200 mA
$
Sol. $J_d=\frac{I}{A}=\frac{6}{16}$
$\therefore$ I through small area $=J_d \times A^{\prime}=\frac{6}{16} \times 3.2=$$
1.2 A=1200 mA
$
Question 44 Marks
Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point $P$ at a distance of 4 cm from wire Y is $x \times 10^{-5} T$. The value of $x$ is _________. Take permeability of free space as $\mu_0=4 \pi \times 10^{-7}$ SI units.
Answer
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Sol.
$\begin{array}{l}B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\ =-\frac{100 \mu_0}{4 \pi} \\ =-100 \times 10^{-7} \\ =-1 \times 10^{-5} T\end{array}$
Sol.
$\begin{array}{l}B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\ =-\frac{100 \mu_0}{4 \pi} \\ =-100 \times 10^{-7} \\ =-1 \times 10^{-5} T\end{array}$
Question 54 Marks
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 ms^{-1}$. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is $x \times 10^4 N / C$. the value of x is _____________.
Take the mass of the proton $=1.6 \times 10^{-27} kg$.
Take the mass of the proton $=1.6 \times 10^{-27} kg$.
Answer
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Sol. For uniform speed $V=\frac{E}{B}$
$
\begin{array}{l}
R=\frac{mV}{eB} \\
=\frac{mV^2}{eE} \\
\Rightarrow E=\frac{mV^2}{eR} \\
=\frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{1.6 \times 10^{-19} \times 2 \times 10^{-2}} \\
=2 \times 10^4 N / C .
\end{array}
$
Sol. For uniform speed $V=\frac{E}{B}$
$
\begin{array}{l}
R=\frac{mV}{eB} \\
=\frac{mV^2}{eE} \\
\Rightarrow E=\frac{mV^2}{eR} \\
=\frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{1.6 \times 10^{-19} \times 2 \times 10^{-2}} \\
=2 \times 10^4 N / C .
\end{array}
$