The design of some physical instrument requires that there be a constant difference in length of 10 cm between an iron rod and a copper cylinder laid side by side at all temperature find their lengths
`(alpha_(Fe)=11 xx 10^(6).^(@)C^(-1),alpha_(Cu)=17 xx 10^(-6) .^(@)C^(-1))`

What should be the length of steel and copper rods at 0^(@)C that the length of steel rod is 5 cm longer than copper at all termperature? Given alpha_(Cu) = 1.7 xx 10^(5) .^(@)C^(-1) and alpha_(steel) = 1.1 xx 10^(5) .^(@)C^(-1) .

The length of the steel rod which would have the same difference in length with a copper rod of length 24cm at all temperatures. (alpha_("copper") = 18 xx 10^(-6) K^(-1) alpha_("steel") = 12 xx 10^(-6) k^(-1)) is -

What must be the lengths of steel and copper rods at 0^(@)C for the difference in their lengths to be 10 cm at any common temperature? (alpha_(steel)=1.2xx10^(-5).^(@)K^(-1)and alpha_("copper")=1.8xx10^(-5).^(@)K^(-1))

In a certain arrangement, two cylinders of brass and steel are laid side by side and there is a constant difference of length between them at any temperature. Calculate the lengths of these cylinders at 0^@C if there is a difference of 5 cm between their lengths at all temperatures. alpha_("brass") = 0.0000019^@C^(-1) alpha_("steel") = 1.2 xx 10^(-6) C^(-1)

Consider a brass rod and a steel rod (80 cm longer than brass rod) at 0^(@)C . It is observed that on increasing temperatures of the two rods by same amount difference in lengths of the two rods does not change. Given that the thermal coefficient of linear expansion for steel and brass are 11xx10^(-6).^(@)C^(-1) and 19xx10^(-6).^(@)C^(-1) respectively. The sum of lengths of the two rods at 0^(@)C is

A carbon filament has resistance of 120Omega at 0^(@)C what must be te resistance of a copper filament connected in series with resistance at all temperature (alpha_("carbon")=(-5xx10^(-4))/( ^(@)C),alpha_("copper")=(4xx10^(-3))/( ^(C))

A bar of iron is 10 cm at 20^(@)C . At 19^(@)C it will be (alpha_(Fe)=11xx10^(-6)//.^(@)C)

Two pendulum clocks, one having an iron pendulum and the other having a brass pendulu, are keeping correct time at 5^(@)C . How much per day will they differ at 25^(@)C ? alpha_("iron")=12xx10^(-5).^(@)C^(-1), alpha_("brass")=18.7xx10^(-6).^(@)C^(-1)