Question

solve this .........

Answer 1

HELLO DEAR,

Let the required tension be T and the tension in the string connecting A and B be T'

now,

★ in "c" ★

mg - T = ma

2g - T = 2a

T = 2(g - a)-----------( 1 )

★ in “B” ★

mg + (T - T") = ma

2g + (T - T") = 2a

2(a - g) + T" = T---------------( 2 )

★ in “A” ★

T" - mg = ma

T" = 2(a + g)----------( 3 )

from----( 1 ) & ---( 2 )

2(g - a) = 2(a - g) + T"

2g - 2a - 2a + 2g = T"

4g - 4a = T"

4g - 4a = 2a + 2g -------- ( 3 )

4g - 2g = 2a + 4a

2g = 6a

a = g/3 [ put in -- -- ( 1 ) ]

T = 2(g - a)

T = 2(g - g/3)

T = 2(3g - g)/3

T = 4g/3

T = 40/3
[ g = 10m/s² ]

T = 13.33 N

--------- or ----------

T = 4g/3

T = 4(9.8)/3
[ g = 9.8m/s² ]

T = 39.2/3

T = 13.0N

HENCE, OPTION ( B ) IS CORRECT


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hoya

Framing Equations for seperate bodies :

$t_{1} - 2g = 2a - > (i)$

$t_{2} + 2g - t_{1} = 2a - > (ii)$

$2g - t_{2} = 2a - > (iii)$

Add all three to get :

$6a = 2g$

$= > a = \frac{g}{3}$

And from here, Put the value of ( a ) in Equation three to get :

$t_{2} = 2g - \frac{2g}{3} = \frac{4g}{3}$

Hence, The tension between the two Blocks B and C is : ( 4g / 3 )

Further, for numeric value, g = 9.8 gives :

$t_{2} = 13N(approx.)$

Note : Acceleration of the system is towards the Block C, downwards as we presumed ^_^