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# Tutor profile: Chirag S.

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Chirag S.
Additive Manufacturing Research & Development Engineer
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## Questions

### Subject:Calculus

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Question:

Evaluate the integral: $$\int_{0}^{1}x\sqrt{3x+1}~dx$$

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Chirag S.

Let $$u=3x+1$$ Differentiate both sides with respect to $$x$$ to get: $$\frac{du}{dx}=3(1)+0$$ $$dx=\frac{du}{3}$$ Changle the limits as shown: When $$x\rightarrow=0$$, $$u\rightarrow=1$$ When $$x\rightarrow=1$$, $$u\rightarrow=4$$ Now, substitute the values: =$$\int_{1}^{4} \frac{u-1}{3}\sqrt{u}~\frac{du}{3}$$ =$$\frac{1}{9}\int_{1}^{4} {(u-1)}\sqrt{u}~{du}$$ =$$\frac{1}{9}\int_{1}^{4} {(u^{\frac{3}{2}}-u^{\frac{1}{2}})}~{du}$$ =$$\frac{1}{9}\left(\frac{2}{5}\times 32-\frac{2}{3}\times 8\right)-\left(\frac{2}{5}-\frac{2}{3}\right)$$ =$$\frac{1}{9}\left(\frac{2}{5}\times 32-\frac{2}{3}\times 8\right)-\left(\frac{2}{5}-\frac{2}{3}\right)$$ On solving the above expression the answer is $$\frac{116}{135}$$

### Subject:Physics

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Question:

Calculate the energy received by a $$2~x~2$$ m solar collector whose normal is inclined at $$45^{\circ}$$ to the sun. The energy loss through the atmosphere is $$50$$ percent and the diffuse radiation is $$20%$$ percent of direct radiation.

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Chirag S.

Given: Surface temperature $$T = 6000 K$$ Distance between earth and sun $$R = 12\times 10^{10}m$$ Diameter on the sun $$D_1 = 1.5\times 10^{9}m$$ Diameter of the earth $$D_2 = 13.2\times 10^{6}m$$ Solution: 1. Energy emitted by sun $$E_b = {\sigma}T^4$$ $$E_b = 5.67\times 10^{-8}\times (6000)^4$$ [$$\sigma = Stefan - Boltzmann~constant = 5.67\times 10^{-8} W/m^2 K^4$$ ] $$E_b = 73.4\times 10^6 W/m^2$$ Area of sun $$A_1 = 4\pi r^2$$ = $$4\pi \times \left(\frac{1.5\times 10^9}{2}\right)^2$$ $$A_1 = 7\times 10^{18}~\text{m}^2$$ Energy emitted by the sun: $$E_b = 73.4\times 10^6 \times 7\times 10^{18}$$ $$E_b = 5.14\times 10^{26}~\text{W}$$ 2) The emission recieved per $$\text{m}^2$$ just outside the earth's atmosphere: The distance between earth and the sun: $$R = 12\times 10^{10}$$ Area $$A = 4\pi R^2$$ =$$4\pi \times (12\times 10^{10})^2$$ $$A = 1.8\times 10^{23} \text{m}^2$$ The radiation received outside the earth atmosphere per $$\text{m}^2$$ =$$\frac{E_b}{A}$$ =$$\frac{5.14\times 10^{26}}{1.8\times 10^{23}}$$ =$$2855.5 ~\frac{\text{W}}{\text{m}^2}$$ 3. Energy received by the earth: Earth area = $$\frac{\pi}{4}(D_2)^2$$ = $$\frac{\pi}{4}\times [13.2x10^6]^2$$ Earth area = $$1.36\times 10^4m^2$$ Energy received by the earth = $$2855.5\times 1.36\times 10^4$$ = $$3.88\times 10^{17} W$$ 4) The energy recieved by a $$2\times 2\times{m}$$ solar collector: Energy loss through the atmosphere is $$50\%$$. So energy reaching the earth: $$100-50=50\%$$ = $$0.5$$ Energy receieved by the earth: =$$0.50\times 2855.5$$ =$$1427.7~\frac{\text{W}}{m^2}$$ Diffuse radiation is $$20\%$$ $$0.20\times 1427.7 = 285.5 \frac{\text{W}}{\text{m}^2}$$ Diffuse radiation = $$285.5 \frac{\text{W}}{\text{m}^2}$$ Total radiation reaching the collection: =$$142.7+285.5$$ =$$1713.2 \frac{\text{W}}{\text{m}^2}$$ Plate area = $$A\times \cos \theta$$ =$$2\times 2\times \cos{45}^{\circ}$$ =$$2.82 \text{m}^2$$ Energy received by the collector: =$$2.82 \times 1713.2$$ =$$4831.2 \text{W}$$

### Subject:Algebra

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Question:

Prove that $$2^n>2n+1$$ for evey $$n>2$$. Explain the steps.

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Chirag S.

Using the principle of mathematical induction: $$1.~Put~n=3$$ $$2^3>2(3)+1$$ $$8>7$$ $$P(n = 3)~is~true$$ $$Let~P(k)~be~true$$ $$2^k>2(2k+1)$$ Now multiply both sides by 2 $$2^k~\times~2>2(2k+1)$$ $$2^{k+1}>2k+2+2k$$ $$2^{k+1}>2k+2+1$$, Since $$(2k>1)$$ $$2^{k+1}>2(k+1)+1$$, We can replace $$2k$$ with $$1$$ $$P(k+1)$$ is true. Hence it is proved.

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