Question

The length ofa metal rod at 0°C is 0.5 m when it is heated, its length increases by 27 mm. The final
temperature ofrodis (a = 90 x10C) 15x then x is ​

Answer 1

Answer:

NSWER

NSWERIn this case, l= 0.5 m

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 m

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10 −6

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10 −6

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10 −6 dt=dl/(lX α) =(3/5) ×10

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10 −6 dt=dl/(lX α) =(3/5) ×10 2

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10 −6 dt=dl/(lX α) =(3/5) ×10 2

NSWERIn this case, l= 0.5 m dl=2.7 mm = 27X 10 −4 mα=90 X10 −6 dt=dl/(lX α) =(3/5) ×10 2 Since the initial temperature is 0 degree, the final temperature will be 60 degree centigrade.