An X-ray tube is operating at 50 kV and 20 mA | vedclass

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(D) The total power supplied is $P = VI = (50 	imes 10^3 V) 	imes (20 	imes 10^{-3} A) = 1000 W$.Since $1\%$ of power is converted into $X$-rays,$9

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All of the above

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(D) The total power supplied is $P = VI = (50 	imes 10^3 V) 	imes (20 	imes 10^{-3} A) = 1000 W$.Since $1\%$ of power is converted into $X$-rays,$99\%$ is converted into heat.Power converted into heat $P_H = 0.99 	imes 1000 W = 990 W$.The rate of rise of temperature is given by $P_H = ms \frac{dT}{dt}$.Substituting the values: $990 = (1.0 kg) 	imes (495 J kg^{-1} {}^\circ C^{-1}) 	imes \frac{dT}{dt}$.$\frac{dT}{dt} = \frac{990}{495} = 2 {}^\circ C/s$. Thus,option $(c)$ is correct.The minimum wavelength is given by $\lambda_{min} = \frac{hc}{eV}$.$\lambda_{min} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{1.6 	imes 10^{-19} 	imes 50 	imes 10^3} \approx 0.248 	imes 10^{-10} m \approx 0.25 	imes 10^{-10} m$. Thus,option $(b)$ is correct.Since the target absorbs a large amount of heat,it must have a high melting point to prevent damage. Thus,option $(a)$ is correct.Therefore,all statements are correct.

An $X$-ray tube is operating at $50 kV$ and $20 mA$. The target material of the tube has a mass of $1.0 kg$ and specific heat $495 J kg^{-1} {}^\circ C^{-1}$. One percent of the supplied electric power is converted into $X$-rays and the entire remaining energy goes into heating the target. Then:

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