Explanation:
$\tau=\text{I}\alpha$ (Magnitude)
For equal torque, we have:
$\text{I}_{\text{A}\propto\text{A}}=\text{I}_{\text{B}\propto\text{B}}$
$\text{I}_{\text{A}}<\text{I}_{\text{B}}$
$\Rightarrow\alpha_{\text{A}}>\alpha_{\text{B}}\ \dots(\text{i})$
Now, $\omega=\alpha\text{t}$
Or, $\frac{\text{v}}{\text{r}}=\alpha\text{t}$
$\text{v}_{\text{A}}>\text{v}_{\text{B}}$ [Using (i)]
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A potential barrier of 0.50 V exists across a P-N junction. If the depletion region is 5.0 
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(a) 1.0 |
(b) 1.0 |
(c) 2.0 |
(d) 2.0 |
A lens which has focal length of 4 cm and refractive index of 1.4 is immersed in a liquid of refractive index 1.6, then the focal length will be
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(a) – 12.8 cm |
(b) 32 cm |
(c) 12.8 cm |
(d) – 32 cm |
Seven capacitors each of capacity 2μF are to be so connected to have a total capacity 
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(a)
|
(b)
|
|
(c)
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(d)
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In the above question, potential difference across the 40Ω resistance will be
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(a) Zero |
(b) 80 V |
(c) 40 V |
(d) 120 V |
The intensity of electric field at a point between the plates of a charged capacitor
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(a) Is directly proportional to the distance between the plates |
|
(b) Is inversely proportional to the distance between the plates |
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(c) Is inversely proportional to the square of the distance between the plates |
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(d) Does not depend upon the distance between the plates |
A point object O is placed on the principal axis of a convex lens of focal length 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. If the eye is placed 60 cm to the right of the lens at a distance h below the principal axis, then the maximum value of h to see the image will be
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(a) 0 |
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(c) 2.5 cm |
(d) 10 cm |
