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Byju's Answer

Standard VI

Mathematics

Disjoint Sets

In above prob...

Question

In above problem choose the correct value(s) of $F$ which the blocks $M$ and $m$ remain stationary with respect to $M_{0}$

(M_{0} + M + m) \frac{g}{μ}

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\frac{m (M_{0} + M + m) g}{M - μ m}

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(M_{0} + M + m) \frac{m g}{M}

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Solution

The correct option is B $\frac{m (M_{0} + M + m) g}{M - μ m}$
$F = (M_{0} + M + m) a$

$M a = T$

$T - μ m a = m g$

$M a - μ m a = m g$

$a = \frac{m g}{M - μ m}$

Therefore,

$F = (M_{0} + M + m) \frac{m g}{M - μ m}$

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Similar questions

Q. Choose the correct value(s) of

F

which the blocks

M

and

m

remain stationary with respect to

M_{0}

Q. Consider a special situtation in which both the faces of the block

M_{0}

are smooth, as shown in adjoining figure. Mark out the correct statement(s)

(A) If

F = 0

, the blocks

M

and

m

cannot remain stationary
(B) For one unique value of

F

, the blocks

M

and

m

remain stationary with respect to block

M_{0}

F

for which blocks

M

and

m

remain stationary with respect to block

M_{0}

(D) Since there is no friction, therefore, blocks

M

and

m

cannot be in equillibrium with respect to block

M_{0}

Q. Consider a special situtation in which both the faces of the block

M_{0}

are smooth, as shown in adjoining figure. Mark out the correct statement(s)

(A) If

F = 0

, the blocks

M

and

m

cannot remain stationary
(B) For one unique value of

F

, the blocks

M

and

m

remain stationary with respect to block

M_{0}

F

for which blocks

M

and

m

remain stationary with respect to block

M_{0}

(D) Since there is no friction, therefore, blocks

M

and

m

cannot be in equillibrium with respect to block

M_{0}

Q. A rope of mass

m_{0}

and length L is suspended vertically. If a mass m is suspended from the bottom of the rope, find the time for a transverse wave to travel the length of the rope.

The bob of a simple pendulum has a mass m and it is executing simple harmonic motion of amplitude A and period T. it collides with a body of mass $m_{0}$ placed at the equilibrium position which sticks to the bob. The time period of the oscillation of the combined masses will be

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In above problem choose the correct value(s) of F which the blocks M and m remain stationary with respect to M0

The correct option is B m(M0+M+m)gM−μmF=(M0+M+m)a Ma=T T−μma=mg Ma−μma=mg a=mgM−μm Therefore, F=(M0+M+m)mgM−μm

In above problem choose the correct value(s) of $F$ which the blocks $M$ and $m$ remain stationary with respect to $M_{0}$

The correct option is B $\frac{m (M_{0} + M + m) g}{M - μ m}$
$F = (M_{0} + M + m) a$

$M a = T$

$T - μ m a = m g$

$M a - μ m a = m g$

$a = \frac{m g}{M - μ m}$

Therefore,

$F = (M_{0} + M + m) \frac{m g}{M - μ m}$