Chapter 11: Problem 37
If the temperature of the sun was to increase from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64
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Chapter 11: Problem 37
If the temperature of the sun was to increase from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64
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Two bodies each having a heat capacity of \(C=500 \mathrm{~J} / \mathrm{K}\) are joined together by a rod of length \(L=40.0 \mathrm{~cm}\), thermal conductivity \(20 \mathrm{~W} / \mathrm{mK}\), and cross-sectional area of \(S=3.00 \mathrm{~cm}^{2}\). The bodies are joined with the help of a thermally insulated rod. The time after which temperature difference diminishes \(\eta=2\) times is (Disregard the heat capacity of the rod.) (A) \(193 \mathrm{~min}\) (B) \(240 \mathrm{~min}\) (C) \(77 \mathrm{~min}\) (D) \(144 \mathrm{~min}\)
A copper sphere is suspended in an evacuated chamber maintained at \(300 \mathrm{~K}\). The sphere is maintained at constant temperature of \(900 \mathrm{~K}\) by heating electrically. A total of \(300 \mathrm{~W}\) electric power is needed to do this. When half of the surface of the copper sphere is completely blackened, \(600 \mathrm{~W}\) is needed to maintain the same temperature of sphere. The emissivity of copper is (A) \(1 / 4\) (B) \(1 / 3\) (C) \(1 / 2\) (D) 1
If the temperature of the sun is increased from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64
If the temperature of the sun is increased from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64
A body cools from \(60^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in \(10 \mathrm{~min}\). If the room temperature is \(25^{\circ} \mathrm{C}\) and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next 10 min will be (A) \(38.5^{\circ} \mathrm{C}\) (B) \(40^{\circ} \mathrm{C}\) (C) \(42.85^{\circ} \mathrm{C}\) (D) \(45^{\circ} \mathrm{C}\)
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