Question

On a hypothetical scale X, the ice point is 40 degree and the steam point is 120 degrees.For another scale Y the ice point and steam point are -30 and 130 respectively.If X-reads degrees .The reading of Y is?

Answer 1

Answer:

It is assumed that all balls are of same mass m and both springs have same force constant k.  

Let x1 be the elongation of spring between A and B. Let x2 be the elongation of spring between B and C.

 

Let aA, aB and aC are acceleration of balls A, B and C respectively

 

(1) equilibrium  

Let T is the tension in the string between Ball-A and support

for Ball-C :-                  k×x2 = m×g  ..............(1)

for Ball-B :-                  m×g + k×x2 = k×x1 .................(2)

                  using eqn.(1),       2×m×g = k×x1 ...................(3)

 

for Ball-A :-               T = m×g + k×x1  .........................(4)

                   using eqn.(3)         T = 3×m×g  ........................(5)

 

(2) When string between ball-A and support is broken

 

for Ball-A :-          m×g + k×x1 = m×aA .............(4)

                     using eqn.(3),  we can rewrite eqn.(4) as ,   3×m×g = m×aA  , aA = 3×g

for Ball-B :-    m×g + k×x2 - k×x1  = m×aB ...................(5)

   using eqn.(2), we know that LHS of  eqn.(5) is zero, hence aB = 0

for Ball-C:-    m×g - k×x2 = m×aC .......................(6)

                      using eqn.(1), we know that LHS of  eqn.(6) is zero, hence aC = 0

 

(3) When spring between ball-A and ball-B is broken

 

Let T1 is the tension in the string between Ball-A and support

 

for Ball-A :-          m×g  =  T1  .............(7)

                      m×g - T1 = m×aA  .............(8)

                      LHS of eqn.(8) is zero because of eqn.(7), hence aA = 0

 

for Ball-B :-    m×g + k×x2   = m×aB ...................(9)

   using eqn.(1), we rewrite eqn.(9) as,   2×m×g    = m×aB

Hence aB = 2×g

 

for Ball-C:-    m×g - k×x2 = m×aC .......................(10)

                      using eqn.(1), we know that LHS of  eqn.(6) is zero, hence aC = 0

 

(3) When spring between ball-B and ball-C is broken

Let T2 is the tension in the string between Ball-A and support

 

for Ball-A :-          m×g + k×x1  =  T2  .............(11)

                      m×g + k×x1 - T2 = m×aA  .............(12)

                      LHS of eqn.(12) is zero because of eqn.(11), hence aA = 0

 

for Ball-B :-    k×x1 - m×g    = m×aB ...................(12)

   using eqn.(3), we rewrite eqn.(9) as,   m×g  = m×aB

Hence aB = g (upwards)

 

for Ball-C:-    m×g  = m×aC .......................(6)

                      hence aC = g  (downwards)