▶️ Answer/Explanation
\[ T^2 \propto a^3 \] \[ \Rightarrow T \propto a^{3/2} \] \[ \Rightarrow 1 \propto \left( 1.5 \times 10^6 \right)^{3/2} \] \[ 2.83 \propto (a)^{3/2} \] \[ \Rightarrow \frac{1}{2.83} = \left( \frac{1.5 \times 10^6}{a} \right)^{3/2} \] \[ \Rightarrow \frac{1}{\left( 2.83 \right)^{2/3}} = \frac{1.5 \times 10^6}{a} \Rightarrow a = 1.5 \times \left( 2.83 \right)^{2/3} \times 10^6 \] \[ = 3 \times 10^6 \, \text{km}\]
Thus, the correct option is (B).
▶️ Answer/Explanation
\( y_1 = \frac{5}{3} \), \( y_2 = \frac{7}{5} \)
\( \frac{y_1}{y_2} = \frac{25}{21} \)
Thus, the correct option is (C).
▶️ Answer/Explanation
Steel is used in the construction of buildings and bridges.
Steel is more elastic and its elastic limit is high.
Thus, the correct option is (A).
▶️ Answer/Explanation
This is called chromatic aberration.
Thus, the correct option is (D).
▶️ Answer/Explanation
For isothermal process: P₃ > P₂ > P₁ at constant volume and graph will be hyperbolic in nature.
Thus, the correct option is (A).
▶️ Answer/Explanation
Emf across its end = \(\frac{1}{2} \text{Bl}^2 \omega\)
Thus, the correct option is (A).
▶️ Answer/Explanation
\( \vec{P} \cdot \vec{Q} = 0 \)
\( (1+2m)(4) + (m)(-2) + (m)(m) = 0 \)
\( 4 + 8m – 2m + m^2 = 0 \)
\( m^2 + 6m + 4 = 0 \Rightarrow (m+2)^2 = 0 \Rightarrow m = 2 \)
Thus, the correct option is (B).
▶️ Answer/Explanation
At Mount Everest g decreases, so time period increases and clock becomes slow (not fast).
Thus, A is not correct but R is correct.
The correct option is (B).
▶️ Answer/Explanation
\( \lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}} \)
Mass: α-particle > proton > electron
Wavelength: \(\lambda_a < \lambda_p < \lambda_e\)
Thus, the correct option is (A).
▶️ Answer/Explanation
When bath switch is open or any one of the switch is
closed bulb will glow. If bath switch is closed; It will not
glow.
Thus, the correct option is (B).
▶️ Answer/Explanation
\( V = \frac{K(\lambda \pi R_1)}{R_1} + \frac{K(\lambda \pi R_2)}{R_2} = 2K\lambda\pi = \frac{\lambda}{2\varepsilon_0} \)
Thus, the correct option is (D).
▶️ Answer/Explanation
Equivalent resistance = 180 \(\Omega\)
Current \( i = \frac{90}{180} = 0.5 \, \text{A} \)
Voltmeter reading = 40 V
Thus, the correct option is (A).
▶️ Answer/Explanation
\( K \Delta x = m\omega^2 (\ell + \Delta x) \)
\( 12.5 \Delta x = 0.2 \times 25 (\ell + \Delta x) \)
\( \frac{\Delta x}{\ell+\Delta x} = \frac{5}{15.5} = \frac{2}{5} \)
\(5\Delta x = 2\ell + 2\Delta \)
\( \frac{\Delta x}{\ell} = 2 : 3 \)
Thus, the correct option is (D).
▶️ Answer/Explanation
\( V = r^a s^b s^c \)
\(\Rightarrow T^{-1} = (L^a)(ML^{-3})^b (MT^{-2})^c \)
\( b + c = 0 \;\;\Rightarrow\;\; b = -c \)
\( a – 3b = 0 \)
\(-1 = 2c \;\;\Rightarrow\;\; c = \dfrac{1}{2} \)
▶️ Answer/Explanation
\( g = \dfrac{GM}{r^2} \quad \text{for outside} \)
\( g = \dfrac{GMr}{R^3} \quad \text{for inside} \)
▶️ Answer/Explanation
For electromagnetic waves: \( E_0 = cB_0 \) and \( c = \frac{\omega}{k} \)
Thus, \( E_0 = \frac{\omega}{k} B_0 \Rightarrow kE_0 = \omega B_0 \)
Thus, the correct option is (D).
▶️ Answer/Explanation
\( \Delta E = 13.6 \left(1 – \frac{1}{16}\right) = 13.6 \times \frac{15}{16} \, \text{eV} \)
\( \lambda = \frac{hc}{\Delta E} = \frac{4 \times 10^{-15} \times 3 \times 10^8}{13.6 \times \frac{15}{16}} \approx 94.1 \, \text{nm} \)
Thus, the correct option is (A).
▶️ Answer/Explanation
Correct matching: A-II, B-I, C-IV, D-III
Thus, the correct option is (C).
▶️ Answer/Explanation
\( n = 70 \, \text{turns/cm} = 7000 \, \text{turns/m} \)
\( B = \mu_0 n I = 4\pi \times 10^{-7} \times 7000 \times 2 = 176 \times 10^{-4} \, \text{T} \)
Thus, the correct option is (A).
▶️ Answer/Explanation
Displacement = \(8\times2-2\times4+4\times4-2\times4 = 32-16 =16\) m,
Distance =\(8\times2+2\times4+4\times4+2\times4 = 48\) m
Ratio = 16 : 48 = 1 : 3
Thus, the correct option is (C).
▶️ Answer/Explanation
3
Long after the switch is closed, the inductors behave like short circuits (zero resistance). The circuit reduces to three resistors in parallel.
Equivalent resistance: \( \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \) ⇒ \( R_{eq} = 4 \Omega \)
Current through battery: \( I = \frac{V}{R_{eq}} = \frac{12}{4} = 3 \text{A} \).
▶️ Answer/Explanation
\( C_o = \dfrac{\varepsilon A}{d} = 15 \, \text{pf} \)
\( C = \dfrac{K \varepsilon_0 A}{2d} = \dfrac{K}{2} \left( \dfrac{\varepsilon_0 A}{d} \right) = \dfrac{105}{4} \, \text{pf} \)
▶️ Answer/Explanation
3696×10-7
▶️ Answer/Explanation
32
Let mass density be ρ.
Mass of original cylinder \(M = ρ × πR²L\)
I1 =\((\frac{1}{2})MR²\) = \( (\frac{1}{2})ρπR²L × R²\) =\( (\frac{1}{2})ρπR⁴L\)
Mass of carved cylinder\( m = ρ × π(R/2)²(L/2) = ρ × π(R²/4)(L/2) = (1/8)ρπR²L = M/8\)
I2 (for carved part) = \( (\frac{1}{2})m(R’)² =(\frac{1}{2}) × (M/8) × (R/2)² = (\frac{1}{2})×(M/8)×(R²/4) = (1/64)MR²\)
But I1 = \((\frac{1}{2}) MR²\)
So, \( \frac{I_1}{I_2} = \frac{(1/2)MR²}{(1/64)MR²} = \frac{64}{2} = 32 \).
▶️ Answer/Explanation
54
Lens maker’s formula: \( \frac{1}{f} = (\mu – 1)\left( \frac{1}{R_1} – \frac{1}{R_2} \right) \)
In air: \( \frac{1}{18} = (1.5 – 1)\left( \frac{1}{R_1} – \frac{1}{R_2} \right) = 0.5K \) ⇒ \( K = \frac{1}{18 \times 0.5} = \frac{1}{9} \)
In water: \( \frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} – 1 \right)K = \left( \frac{1.5}{4/3} – 1 \right) \times \frac{1}{9} = \left( \frac{4.5}{4} – 1 \right) \times \frac{1}{9} = \left( \frac{0.5}{4} \right) \times \frac{1}{9} = \frac{1}{8} \times \frac{1}{9} = \frac{1}{72} \)
So, fw = 72 cm.
Change in focal length = fw – fair = 72 – 18 = 54 cm.
▶️ Answer/Explanation
100
\(\vec{f} = (ti + 3t^2 j)\,\text{N}\)
\(\vec{a} = ti + 3t^2 j\)
\(\vec{v} = \dfrac{t^2}{2} i + \dfrac{3t^3}{3} j = \dfrac{t^2}{2} i + t^3 j\)
\(p = \vec{f} \cdot \vec{v} = \dfrac{t^3}{3} + 3t^5 = \dfrac{8}{2} + 3(2)^5 = 4 + 96 = 100\)
▶️ Answer/Explanation
44%
▶️ Answer/Explanation
\(\frac{9}{130}\)
▶️ Answer/Explanation
6
Molar mass of \(^{240}X\) = 240 g/mol.
Number of moles in 120 g = 120 / 240 = 0.5 mol.
Number of atoms N = 0.5 × NA = 0.5 × 6.022 × 1023 = 3.011 × 1023
Energy released = N × 200 MeV = 3.011 × 1023 × 200 = 6.022 × 1025 MeV ≈ 6 × 1025 MeV.
So, the value is 6.
▶️ Answer/Explanation
1
Time period T = \( 2\pi\sqrt{\frac{m}{k}} \)
Case 1: 1 = \( 2\pi\sqrt{\frac{m}{k}} \) ⇒ \( \frac{m}{k} = \frac{1}{4\pi^2} \) …(1)
Case 2: 2 = \( 2\pi\sqrt{\frac{m+3}{k}} \) ⇒ \( \frac{m+3}{k} = \frac{4}{4\pi^2} = \frac{1}{\pi^2} \) …(2)
Divide (2) by (1): \( \frac{m+3}{m} = \frac{1/\pi^2}{1/(4\pi^2)} = 4 \) ⇒ m + 3 = 4m ⇒ 3m = 3 ⇒ m = 1 kg.