Question

The system shown below is initially in equilibrium. Masses of the blocks A, B, C, D and E are respectively 3m, 3m, 2m, 2m and 2m, Match the conditions in column-I with the effect in

Column-II.

COLUMN 1

(p) After spring 2 is cut, tension in string AB

(q) After spring2 is cut, tension in string CD

(r) After string between C and pulley is cut, tension in string AB

(s) After string between C and pulley is cut, tension in string CD

COLUMN 2

(a) increases

(b) decreases

(c) remains constant

(d) Zero

• A. p - c, q - b, r - d, s - b
• B. p - d, q - b, r - b, s - b
• C. p - d, q - d, r - d, s - d
• D. p - c, q - b, r - c, s - b

Answer 1

The correct option is D

p - c, q - b, r - c, s - b

Let's first analyse the equilibrium condition.

$T_{AB}=3mg$ -----------(i)

$K_1{X}_1$ = 3mg + $T_{AB}$ From Eq (I)

$K_1{X}_1$ = 6mg (II)

$K_1{X}_1$ = $T_{CD}+2mg$ From (ii)

$T_{CD}=4mg$ -----------(iii)

$T_{CD}=K_2{X}_2+2mg$ from (iii)

From (ii)

$K_2{X}_2=2mg$ -----------(iv)

$K_2{X}_2=2mg$ well we got that before

Now lets take first problem

$T_{AB}$ just after cutting spring 2

Hmmm............

Let's first see what I can be sure of spring 2's $k_2{x}_2$ will be directly 0 but spring 1 will have $k_1{x}_1$ which won't immediately become 0 and since its asked just after cutting so spring 2 force will be $\simeq$ 6mg. Fair assumption

Free body drawing of B

and we know B won't start falling immediately after cutting the spring as spring 1 is still stretched.

So $T_{AB}$ = 3mg as before so $T_{AB}$ remains constant. (p - c)

(q) okay now lets focus on the string between C - D

Lets assume its new tension is $T_1$

The C - D system with both blocks will go up as the spring holding them down is now cut.

6mg - T - 2mg = 2ma

4mg - T = 2 ma

T - 2mg = 2 ma

solving above 2 eq

we get T = 3mg

initially $T_{CD}$ = 4mg

Now $T_{CD}$ = 3mg

So reduced.

(q) - (b)

(r) Just after you cut the string the spring is still elongated and will have $k_1{x}_1$

This means block B won't move so 6mg = T + 3mg

T = 3mg remains constant (r - c)

(s) Just after string between C and pulley cut blocks C & D won't have any force to stop them from falling. They are being pulled down by the spring. So an acceleration there common for both C & D

Let's assume there is some tension in CD

_To

$T_0+2mg=2ma$

2mg + $k_2{x}_2$ - $T_0$ = 2ma

4mg - $T_0$ = 2ma

2$T_0$ - 2mg = 0

$T_0$ = mg

Initially $T_{CD}$ = 2mg

Now $T_{CD}$ = mg

So Tension has reduced.