In the circuit element given here, if the potential at B, VB =0, then the potentials A and D are given as
a) VA = – 1.5 V, VD = + 2 V b) VA = – 1.5 V, VD = + 0.5... View Article
a) VA = – 1.5 V, VD = + 2 V b) VA = – 1.5 V, VD = + 0.5... View Article
a) 1.6 x 10-3 rad b) 1.6 x 10-4 rad c) 3.2 x 10-3 rad d) 3.2 x 10-4 rad... View Article
a) 2 x 10-5 H b) 3 x 10-5 H c) 4 x 10-5 H d) 6 x 10-5 H... View Article
a) 2.4 m/s2 b) 9.4 m/s2 c) 6.9 m/s2 d) 4.9 m/s2 Answer: d) 4.9 m/s2 Solution: Fnet = 3T =... View Article
a) 7 b) 0.7 c) 70 d) 0.07 Answer: b) 0.7 N Solution: Given that, layer thickness dy=1mm=10−3m Area A=10−2m2 Coefficient... View Article
a) 0.50 T b) 0.02 T c) 10-3 T d) 0.1 T e) 0.01 T Answer: e) 0.01 T Solution: Given,... View Article
a) l,d b) 2l,d c) (1/2), 2d d) 2l, d/2 Answer: c) (1/2), 2d Solution: Resistance , R = ρ (L/A) = ρ... View Article
Equipotential surfaces a) will always be equally spaced b) are closer in regions of large electric fields compared to regions... View Article
(a) 4i – 13j+ 6k (b) -18i-13j+2k (c) 6i+2J-3k (d) 6i-2j+8k Answer: (b) -18i-13j+2k Solution: v = ω x r = (3i-4j+k) x... View Article
a) \(\begin{array}{l}\sqrt{\frac{4\pi \varepsilon _{0}G}{M_{e}M_{m}}}\end{array} \) b) \(\begin{array}{l}\sqrt{\frac{GMm}{k}}\end{array} \) c) \(\begin{array}{l}\sqrt{\frac{GM_{e}M_{m}}{4\pi \epsilon _{0}}}\end{array} \) d) \(\begin{array}{l}\sqrt{\frac{M_{e}M_{m}}{4\pi \epsilon _{0}G}}\end{array} \) Answer: b) \(\begin{array}{l}\sqrt{\frac{GMm}{k}}\end{array} \) Solution: Gravitational Attraction F=... View Article
a) \(\begin{array}{l}\frac{a_{1}}{a_{2}}=\frac{s_{1}}{s_{2}}=\frac{t_{1}}{t_{2}}\end{array} \) b) \(\begin{array}{l}\frac{a_{1}}{a_{2}}=\frac{s_{2}}{s_{1}}=\frac{t_{1}}{t_{2}}\end{array} \) c) \(\begin{array}{l}\frac{a_{1}}{a_{2}}=\frac{s_{1}}{s_{2}}=\frac{t_{2}}{t_{2}}\end{array} \) d) \(\begin{array}{l}\frac{a_{1}}{a_{2}}=\frac{s_{2}}{s_{1}}=\frac{t_{2}}{t_{1}}\end{array} \) Solution: d) \(\begin{array}{l}\frac{a_{1}}{a_{2}}=\frac{s_{2}}{s_{1}}=\frac{t_{2}}{t_{1}}\end{array} \)
Solution: Total end correction of open pipe \(\begin{array}{l}x_{t}=0.8 cm\end{array} \) Open pipe has end correction on both sides. End correction... View Article
a)The bar first moves downwards and then upwards. b) The bar stops for a moment after 2.0 s from the... View Article
Solution: The equatorial magnetic field is given by the formula: \(\begin{array}{l}B=\frac{\mu_{0} }{4\pi}.\frac{m}{r^{3}}\end{array} \) \(\begin{array}{l}B=0.4*10^{-4}T\end{array} \) Radius of earth r=\(\begin{array}{l} 6.4*10^{6}m\end{array}... View Article
a) \(\begin{array}{l}\vec{v}=\frac{E}{B}\hat{i}+a\hat{j}\end{array} \) b) \(\begin{array}{l}\vec{v}=\frac{E}{B}\hat{i}+b\hat{k}\end{array} \) c) \(\begin{array}{l}\vec{v}=\frac{E}{B}\hat{i}+c\hat{i}\end{array} \) d) \(\begin{array}{l}\vec{v}=\frac{E}{B}\hat{i}\end{array} \) Solution: b) \(\begin{array}{l}\vec{v}=\frac{E}{B}\hat{i}+b\hat{k}\end{array} \)
Solution: Position is given as x=a+bt^{2} x=a+bt^{2} = 8.5+2.5t^{2} ————–(1) Substituting t=2s in equation (1) =18.5 m Substituting t=4s in... View Article
a) 2 b) 4 c) 5 d) 10 Solution:c) 5
Solution: The direction of concentric circular field lines of magnetic field will be in anti-clockwise direction and along the path... View Article
Solution: \(\begin{array}{l}\sqrt{\frac{1}{T}\int_{0}^{T}(10)^{2}dt}\end{array} \) \(\begin{array}{l}\sqrt{\frac{1}{T}(10)^{2}Tdt}\end{array} \) \(\begin{array}{l}\sqrt{(10)^{2}dt} = 10V\end{array} \)
a) b) c) d) Solution: Option a). Explanation: Here, the presence of a non uniform field is seen due to... View Article