JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.
Q131. A room at 20°C is heated by a heater of resistance 20 ohm connected to 200 V mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area 1 m2 and thickness 0.2 cm. Calculate the temperature outside. Thermal conductivity of glass is 0.2 cal/m C° s and mechanical equivalent of heat is 4.2 J/cal
Q132. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T_1 and T_2 respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of ?
Q133. The graph of elongation of rod of a substance A with temperature rise is shown in Figure. A liquid B contained in a cylindrical vessel made up of substance A, graduated in millitres at 0°C is heated gradually. The readings of the liquid level in the vessel corresponding to different temperatures are shown in the figure. The real volume expansivity of liquid is
Q134. It takes 10 min for an electric kettle to heat a certain quantity of water from 0°C to 100°C. It takes 54 min to convert this water at 100°C into steam. Then latent heat of steam is
Q135. A clock with a metal pendulum beating seconds keeps correct time at 0°C. If it loses 12.5 s a day at 25°C, the coefficient of linear expansion of metal of pendulum is
Solution
135 (a) Loss of time due to heating a pendulum is given as ∆T=1/2 α∆θT ⇒12.5=1/2×α×(25-0)℃×86400s ⇒α=1/86400/℃
135 (a) Loss of time due to heating a pendulum is given as ∆T=1/2 α∆θT ⇒12.5=1/2×α×(25-0)℃×86400s ⇒α=1/86400/℃
Q136. The loss in weight of a solid when immersed in a liquid at 0°C is W_0 and at t°C is W. If cubical coefficients of expansion of the solid and the liquid are γ_s and γ_L, respectively, then W is equal to
Q137. A uniform solid brass sphere is rotating with angular speed ω_0 about a diameter. If its temperature is now increased by 100°C, what will be its new angular speed. (Given α_B=2.0×10(-5) per °C)
Solution
Q138. A steel tape is placed around the earth at the equator. When the temperature is 0°C neglecting the expansion of the earth, the clearance between the tape and the ground if the temperature of the tape rises to 30°C, is nearly (α_steel=11×10(-6)/K)
Q139. Water falls from a height 500 m. The rise in temperature of water at bottom if whole of energy remains in water, will be (specific heat of water is c=4.2 kJ kg(-1))