Question

A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is $fraction numerator straight m l squared over denominator 3 end fraction$ )

$fraction numerator 3 straight g over denominator 2 straight l end fraction$
$fraction numerator 2 straight l over denominator 3 straight g end fraction$
$fraction numerator 3 straight g over denominator 2 straight l squared end fraction$
$mg straight l over 2$

Solution

A.

fraction numerator 3 straight g over denominator 2 straight l end fraction

The moment of inertia of the uniform rod about an axis through the end and perpendicular to length is
$straight I space equals space fraction numerator straight m l squared over denominator 3 end fraction$
where m is mass of rod and l its length.
Torque $left parenthesis straight t space equals space Iα right parenthesis$ acting on centre of gravity of rod is given by
$straight t equals mg l over 2$
or $Iα space equals space mg l over 2$
$or space space fraction numerator straight m l squared over denominator 3 end fraction straight alpha space equals space mg l over 2 therefore space space space space straight alpha space equals fraction numerator 3 straight g over denominator 2 l end fraction$

Question

Wired Faculty App

Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]

350 Hz

311 Hz

411 Hz

448 Hz

Solution

D.

448 Hz

straight f subscript straight A space equals space straight f open square brackets fraction numerator straight v plus straight v subscript 0 over denominator straight v minus straight v subscript straight s end fraction close square brackets space equals space 400 space open square brackets fraction numerator 340 space plus 16.5 over denominator 340 minus 22 end fraction close square brackets straight f subscript straight A space equals space 448 space Hz

Question

Wired Faculty App

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I’ along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire ‘B’ is given by

$fraction numerator straight mu subscript straight o straight I over denominator 2 πd end fraction$
$fraction numerator 2 straight mu subscript straight o straight I over denominator πd end fraction$
$fraction numerator square root of straight mu subscript straight o straight I end root over denominator πd end fraction$
$fraction numerator straight mu subscript 0 straight I squared over denominator square root of 2 πd end fraction$

Solution

D.

fraction numerator straight mu subscript 0 straight I squared over denominator square root of 2 πd end fraction

Force between BC and AB will be same in magnitude.

$straight F subscript BC space equals straight F subscript BA space equals space fraction numerator straight mu subscript 0 straight I squared over denominator 2 πd end fraction straight F space equals space square root of 2 straight F subscript BC space equals space square root of 2 fraction numerator straight mu subscript 0 straight I squared over denominator 2 πd end fraction straight F space equals space fraction numerator straight mu subscript 0 straight I squared over denominator square root of 2 πd end root end fraction$

Question

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A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

3D/2

D

5D/4

7D/5

Solution

C.

5D/4

As track is frictionless, so total mechanical energy will remain constant,

$i . e ., 0 + m g h = \frac{1}{2} m v_{L}^{2} + 0 u \sin g v^{2} - u^{2} = 2 g h, h = \frac{v_{L}^{2}}{2 g} (∵ u = 0) F o r c o m p l e t i n g t h e v e r t i c a l c i r c l e V_{L} \geq \sqrt{5 g R} o r, h = \frac{5 g R}{2 g} = \frac{5}{2} R = \frac{5}{4} D$

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Laws of Motion

Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]

350 Hz

311 Hz

411 Hz

448 Hz

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

3D/2

D

5D/4

7D/5

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

WhatsApp Group

Download Android App

Ncert Book Download

Laws of Motion

A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: (Moment of inertia of rod about A is )

Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s] 350 Hz 311 Hz 411 Hz 448 Hz

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I’ along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire ‘B’ is given by

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to 3D/2 D 5D/4 7D/5

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]

350 Hz

311 Hz

411 Hz

448 Hz

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

3D/2

D

5D/4

7D/5