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Asked by meenatoofansingh3 | 15 Apr, 2021, 06:16: PM

Expert Answer

Qn.(6)

Let us consider 6N force is acting at D so that BD = x and AD ( 3 - x )

At equilibrium, sum of moments of forces is zero.

Let us take moments of forces about the point D.

At equilibrium, we have , 2 ( 3 - x ) = 4 x

Frome above expression, we get x = 1

Answer :- 6N force acting downward at a point D so that BD = 1 m

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Figure shows the tension force T acting along the string .

At the disk of mass M , tension force T is giving torque to rotate the disc with angular acceleration α

Hence we have, T × R = I α .....................(1)

where R is radius of Disc , I = (1/2) M R²is the moment of inertia and M is mass of disc

Angular acceleration α = a / R ,

where a is linear accelertion at rim of disc that is same as acceleration of block of mass m

Hence we rewrite eqn.(1) as , T × R = (1/2) M R² × ( a / R )

Hence we get , T = (1/2) M a ...........................(2)

At the block of mass m , by applying Newton's second law , we get

( m g ) - T = m a ...........................(3)

By adding eqn.(2) and (3) , we get acceleration as

begin mathsize 14px style a space equals space fraction numerator m over denominator open parentheses m space plus space M over 2 close parentheses end fraction space g end style

..............................(4)

Hence , using eqn.(2) and (4) , we get tension T as

begin mathsize 14px style T space equals space 1 half M cross times fraction numerator m over denominator open parentheses m space plus space begin display style M over 2 end style close parentheses end fraction space g end style

.......................(5)

Let us substitute values M = 2.4 kg , m = 1.2 kg , g = 9.8 m/s²

We get tension T = (1/2) × 2.4 × [ 1.2 / ( 1.2 + 1.2 ) ] × 9.8 = 5.88 N

Answered by Thiyagarajan K | 15 Apr, 2021, 09:44: PM