Physics 2005 Jamb Examination

Formula: \(Q = mL\). So, \(L = Q/m\).

Given \(Q = 10^6\,\text{J}, m = 2\,\text{kg}\):

\[L = \dfrac{10^6}{2} = 5.0 \times 10^5 \, \text{J kg}^{-1}.\]

Correct option: D.

For capacitors in series: \(\tfrac{1}{C_{eq}} = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6} = 1\). So \(C_{eq} = 1\,\mu\text{F}.\)

Total charge: \(Q = C_{eq}V = 1 \times 12 = 12 \, \mu\text{C}.\)

Charge on each capacitor is the same. Voltage across 6 μF: \(V = Q/C = 12/6 = 2\,\text{V}.\)

Correct option: C.

Coulomb’s law: \(F \propto 1/r^2\). If distance \(r\) becomes 4 times, force becomes \(1/4^2 = 1/16\) of its former value.

Correct option: B.

Velocity of sound in air: \(v \propto \sqrt{T}\) where \(T\) is absolute temperature.

To double \(v\), need \(\sqrt{T_{new}/T_{old}} = 2 \implies T_{new}/T_{old} = 4.\)

So the temperature must be quadrupled. Correct option: A.

Force \(F \propto 1/r^2\). If distance is halved, \(F_{new} = F \times (1/(1/2)^2) = 4F.\)

So new force = \(200 \times 4 = 800\,N.\)

Correct option: D.

For inductors in series, \(L_{eq} = L_1 + L_2 = 3 + 6 = 9\,H.\)

Correct option: A.

Extension due to 40 N: \(36 - L_0\).
Extension due to 60 N: \(45 - L_0\).

For Hooke's law F = Ke

\[ \implies \dfrac{F}{e} = K\]

\[\dfrac{f_1}{e_1} = \dfrac{f_2}{e_2}\]

\[\text{Let the original length} = t_0\]

therefore 

\[e_1 = (36 - t_0)cm; e_2 = (45 - t_0)cm\]

\[f_1 = 40N, f_2 = 60N\]

\[ \dfrac{40}{36 - t_0} = \dfrac{60}{45 - t_0}\]

\[ \implies 40(45 - t_0) = 60(36 - t_0)\]

Therefore \(1800 - 40t_0 = 2160 - 60t_0\)

\(60t_0 - 40t_0 = 2160 - 1800\)
\(20t_0 = 360\)
\(t_0 = 18cm\)

Option D

is the graph of stopping potential Vo against the frequency of the incident light I. The straight line graph cuts the intercept I = threshold frequency. As the frequency of the incident light increases above this point(threshold frequency), the stopping potential corresponding to the kinetic energy of the photoelectron also increases.

Einstein’s relation: $$E = m c^2$$ $$ m = 0.5 \, \text{kg}, \quad c = 3 \times 10^8 \, \text{m/s} $$ $$ E = 0.5 \times (3 \times 10^8)^2 = 0.5 \times 9 \times 10^{16} = 4.5 \times 10^{16} \, \text{J}. $$

Correct option: D.

Equation of motion: \(v^2 = u^2 + 2as\).

Here, \(v = 0, u = 30, s = 50, a = ?\).

\(0 = 30^2 + 2a(50) ⇒ 0 = 900 + 100a ⇒ a = -9\,ms^{-2}.\)

Deceleration = 9 ms−2.

Correct option: C.