473 degree C. 1.200 degree C. A steel wire, 150 m long at 10 degree C, has a coe

Question

473 degree C. 1.200 degree C. A steel wire, 150 m long at 10 degree C, has a coefficient of linear expansion of 11 times 10 /C . Give its change in length as the temperature changes from 10 degree C to 45 degree C. 0.65 cm 1.8 cm 5.8 cm 12 cm At 20 degree C an aluminum ring has an inner diameter of 5.000 cm, and a brass rod has a diameter of 5.050 cm. Keeping the brass rod at 20 degree C, which of the following temperatures of the ring will allow the ring to just slip over the brass rod? (alpha = 2.4 times 105/C4, alpha = 1.9 times 105/C4) 111 degree C 236 degree C 384 degree C 437 degree C

Explanation / Answer

this is easy enough.. here you go buddy initial temperature t1 = 20^ 0C final temperature at which the ring just slip over the brassrod = t2 = ? D1 = 5 cm D2 = 5.05 cm aAl= 2.4 x10^-5
D2- D1 = D1*aAl*(t2-t1) 0.05 = 5*2.4*10^-5*(t2-20)
t2-20=0.05/( 12*10^-5) t2-20 = 416.66 t2 = 436.66^0C      so, if you round that your answer is D please rate:- LiFeSaver
initial temperature t1 = 20^ 0C final temperature at which the ring just slip over the brassrod = t2 = ? D1 = 5 cm D2 = 5.05 cm aAl= 2.4 x10^-5
D2- D1 = D1*aAl*(t2-t1) 0.05 = 5*2.4*10^-5*(t2-20)
t2-20=0.05/( 12*10^-5) t2-20 = 416.66 t2 = 436.66^0C      so, if you round that your answer is D please rate:- LiFeSaver
initial temperature t1 = 20^ 0C final temperature at which the ring just slip over the brassrod = t2 = ? D1 = 5 cm D2 = 5.05 cm aAl= 2.4 x10^-5
D2- D1 = D1*aAl*(t2-t1) 0.05 = 5*2.4*10^-5*(t2-20)
t2-20=0.05/( 12*10^-5) t2-20 = 416.66 t2 = 436.66^0C      so, if you round that your answer is D please rate:- LiFeSaver
initial temperature t1 = 20^ 0C final temperature at which the ring just slip over the brassrod = t2 = ? D1 = 5 cm D2 = 5.05 cm aAl= 2.4 x10^-5
D2- D1 = D1*aAl*(t2-t1) 0.05 = 5*2.4*10^-5*(t2-20)
t2-20=0.05/( 12*10^-5) t2-20 = 416.66 t2 = 436.66^0C      so, if you round that your answer is D please rate:- LiFeSaver

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