Physics Test - 26

Given:

\(f _{1}=50\) cm (convex lens)

\(f _{2}=-25\) cm (concave lens)

We know that:

The focal length of the combination:

\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

\(\frac{1}{f}=\frac{1}{50}+\frac{-1}{25}\)

\(f =-50\) cm

Given:

\(L=40\) cm \(=0.4\) m

\(W =10\) cm \(=0.1\) m

\(N =10\) turns

\(I =16\) A

\(\theta=60^{\circ}\)

\(B =0.60\) T

Where, \(N=\) number of turns in the coil, I = current in the loop, \(A=\) area enclosed by the loop, \(L=\) length of the loop, \(W=\) width of the loop, \(B=\) magnetic field intensity, and \(\theta=\) angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

The area enclosed by the loop is given as,

\(A = L \times W\)

\(\Rightarrow A =0.4 \times 0.1\)

\(\Rightarrow A =0.04\) m\(^{2}\)

So, the torque on the current-carrying rectangular loop is given as,

\(\tau =IAB \sin \theta\)

\(\Rightarrow \tau =16 \times 0.04 \times 0.6 \times \sin 60\)

\(\Rightarrow \tau=\frac{16 \times 4 \times 6 \times \sqrt{3}}{10 \times 10 \times 2}\)

\(\Rightarrow \tau=1.92 \sqrt{3}\) N-m

Given:

\(N_{a}=5 \times 10^{15} \mathrm{~cm}^{-3}\)

\(\mathrm{N}_{\mathrm{d}}=1 \times 10^{15} \mathrm{~cm}^{-3}\)

\(V_{T}=0.0259\)

Intrinsic carrier concentration \(\mathrm{n}_{\mathrm{i}}=1.5 \times 10^{10} \mathrm{~cm}^{-3}\).

The built-in voltage will be:

\(V_{b i}=V_{T} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}}\right)\)

Now putting the given values in the above equation,

\(V_{b i}=0.0259 \ln \left(\frac{\left(5 \times 10^{15}\right)\left(10^{15}\right)}{\left(1.5 \times 10^{10}\right)^{2}}\right)\)

\(\Rightarrow V_{b i}=0.617 \mathrm{~V}\)

The given applied voltage is:

\(V_{a}=\frac{1}{2} V_{b i}\)

\(\Rightarrow V_{a}=\frac{1}{2} \times 0.617 V\)

\(V_{a}=0.308 \mathrm{~V}\)

We know that:

(A) Root mean square speed of gas molecule\(=\sqrt{\frac{3 R T}{M}}\)

(B) Pressure exerted by ideal gas \(=\frac{1}{3} n m \bar{v}^{2}\)

(C) Average kinetic energy of a molecule \(=\frac{3}{2} k_{B} T\)

(D) Total internal energy of a gas, \((U)=\frac{1}{2} n f R T\)

Here, \(n=1\)

\(f=5\)

\(U=\frac{5}{2} R T\)

Therefore, the correct match is as follows:

(A) - (Q), (B) - (P), (C) - (S), (D) - (R)

Let, 

Effective capacitance = C 

Then:

\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)

\(=\frac{1}{8}+\frac{1}{16}+\frac{1}{24}\)

\(=\frac{6+3+2}{48}\)

\(C=\frac{48}{11} \mu F\)

Potential difference, \(V=110 {~V}\)

We know that:

Charge, \(Q=C V\)

\(Q=\frac{48}{11} \times 10^{-6} \times 110\)

\(=480 \times 10^{-6}\)

\(=48 \times 10^{-5}\)

Since the capacitors are in series each capacitor has same charge i.e., \(48 \times 10^{-5} {C}\).

Given:

Output power \(\mathrm{P}=100 \mathrm{~W}\)

Voltage across primary \(\mathrm{V}_{\mathrm{p}}=220 \mathrm{~V}\)

Current in the primary \(\mathrm{I}_{\mathrm{p}}=0.5 \mathrm{~A}\)

Efficiency of a transformer:

\(\eta=\frac{\text { output power }}{\text { input power }} \times 100\) \(=\frac{P}{V_{p} I_{p}} \times 100\)

\(=\frac{100}{220 \times 0.5} \times 100=90 \%\)

The resistance of the wire is directly proportional to the length of the wire and here we are cutting the wire into n equal parts, so the new resistance will be:

\(R _{1}= R _{2}= R _{3}=\ldots \ldots \ldots \ldots R _{ n }= \frac{R}{n}\)

Now, the resistances are placed in parallel combination, therefore, the equivalent resistance will be

\(\frac{1}{R_{ eq }}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\ldots \ldots \ldots \ldots \frac{1}{R_{n}}\)

\(\frac{1}{R_{e q}}=n\left[\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \ldots \ldots \ldots \ldots +\frac{1}{R}\right]=n\left[\frac{n}{R}\right]\)

\(R_{e q}=\frac{R}{n^{2}}\)

Given:

\(B=7 \pi \times 10^{-5}\) Wb/m\(^{2}\) and loop of radius \(=5\) cm \(=5 \times 10^{-2}\) m

The magnetic field at the center of the circular coil is given by,

\(B=\frac{\mu_{o}}{2} \frac{I}{r}\)

\(\Rightarrow I=\frac{2 B r}{\mu_{o}}\)

Where, \(B =\) strength of the magnetic field, \(I =\) current, \(r =\) radius or distance

Substitute the value of \(B, r\), and \(\mu_{0}\) in the above equation, we get,

\(I=\frac{2 \times 7 \pi \times 10^{-5} \times 5 \times 10^{-2}}{4 \pi \times 10^{-7}}=17.5\) A

In a refracting telescope, the objective lens is a convex lens.

• The convex lens gathers the light and converges at the focal point.
• The convex lens in the eyepiece then takes the light from the focal point and makes them parallel. This causes magnification of the point image formed at the focal point.
• Thus, we are able to see magnified images of distant objects.