Question

24. A light string passes over a frictionless pulley. To one of its ends a mass of 6 kg is attached. To its other end a mass of 10 kg is attached. The tension in the thread will be (a) 24.5 N (b) 2.45 N 6 kg 10 kg (C) 79 N (d) 73.5 N​

Answer 1

Answer:

(C) 73.5 N

Explanation:

t - 6g = 6a  -------->   (i)

10g - t = 10a  ------>  (ii)

Adding equation (i) and (ii)

10g - 6g  = 10a + 6a

a = 4g /16 = g/4

a = g / 4

T = 10g - 10a

T = 10 ( g - g/4 )

T = 30g / 4

T  =7.5  x  9.8

T = 73.5N

Hope it helps u !!

Answer 2

t - 6g = 6a _____eq(1)

10g - t = 10a _____eq(2)

add (1),(2)

⇒ 10g - 6g = 6a + 10a

⇒ 4g = 16a

⇒ g = 4a

⇒ a = g/4 ____eq(3)

sub eq(3) in eq(2),

t = 10(g + a)

⇒ t = 10(g - g/4)

⇒ t = 10(3g/4)

⇒ t = 30g/4

⇒ t = 30/4 × 9.8

⇒ t = 73.5N