A beaker contains 200, g of water. The heat c | vedclass

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(B) According to the principle of calorimetry,the heat lost by the hot water is equal to the heat gained by the cold water and the beaker.Let the fina

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$68$

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(B) According to the principle of calorimetry,the heat lost by the hot water is equal to the heat gained by the cold water and the beaker.Let the final equilibrium temperature be $	heta$.Heat lost by hot water = $m_{hot} \cdot c \cdot (T_{hot} - 	heta) = 440 \cdot c \cdot (92 - 	heta)$.Heat gained by cold water = $m_{cold} \cdot c \cdot (	heta - T_{cold}) = 200 \cdot c \cdot (	heta - 20)$.Heat gained by the beaker = $m_{eq} \cdot c \cdot (	heta - T_{cold}) = 20 \cdot c \cdot (	heta - 20)$.Equating heat lost and heat gained:$440(92 - 	heta) = 200(	heta - 20) + 20(	heta - 20)$$440(92 - 	heta) = 220(	heta - 20)$$2(92 - 	heta) = 	heta - 20$$184 - 2	heta = 	heta - 20$$3	heta = 204$$	heta = 68^\circ C$.

List-$I$	List-$II$
$(A)$ Magnetic field intensity	$(i)$ $Wb$
$(B)$ Magnetic flux	(ii) $Wb \cdot m^{-2}$
$(C)$ Magnetic pole strength	(iii) $A \cdot m$
$(D)$ Magnetic induction	(iv) $A \cdot m^{-1}$

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