Find the The Slopes of the tangent and the normal to the following curves at the indicated points:
Question: Find the The Slopes of the tangent and the normal to the following curves at the indicated points: $y=\sqrt{x^{3}}$ at $x=4$ Solution: Given: $y=\sqrt{x^{3}}$ at $x=4$ First, we have to find $\frac{d y}{d x}$ of given function, $f(x), i . e$, to find the derivative of $f(x)$ $y=\sqrt{x^{3}}$ $\therefore \sqrt[n]{x}=x^{\frac{1}{n}}$ $\Rightarrow y=\left(x^{3}\right)^{\frac{1}{2}}$ $\Rightarrow y=(x)^{\frac{3}{2}}$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\...
Read More →In a refrigerator, one removes heat from a lower
Question: In a refrigerator, one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from -3oC to 27oC, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine. Solution: Efficiency of the Carnot engine is $\eta=1-\frac{T_{2}}{T_{1}}$ Where T1= 300 K T2= 270 K It...
Read More →Consider a cycle tyre being filled
Question: Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre and at each stroke of the pump ∆V of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P1to P2? Solution: Following is the equation before and after the stroke: $P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}$ $P(V+\Delta V)^{\gamma}=(P+\Delta P) V^{\gamma} \Rightarrow P V^{\gamma}\left(1+\frac{\Delta V}{V}\right)^{\gamma}=P\left(1+\frac{\D...
Read More →A person of mass 60 kg wants
Question: A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilocalories, how many times must he go up and down to reduce his weight by 5 kg? Solution: Given, Height of the stairs, h = 10 m Work done to burn 5 kg of fat = (5)(7000 103)(4.2) = 147 106J Work done towards burning of fat in one trip = mgh + 1/2 mgh = 3/2 mgh = (3/2)(60)(10)(10) = 9 103J Theref...
Read More →Consider a Carnot’s cycle operating between
Question: Consider a Carnots cycle operating between T1= 500K and T2= 300K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs. Solution: Given, Temperature of the source, T1 = 500 K Temperature of the sink, T2 = 300 K Work done per cycle, W = 1 kJ = 1000 J Heat transferred to the engine per cycle, Q1 =? Efficiency of a Carnot engine $\eta=1-\frac{T_{2}}{T_{2}}=1-\frac{300}{500}=\frac{2}{5}$ $\eta=\frac{W}{Q_{1}} \Rightarrow Q_{1}=\frac{W}{\eta}...
Read More →Air pressure in a car tyre increases
Question: Air pressure in a car tyre increases during driving. Explain. Solution: Air pressure in a car tyre increases during driving while the volume remains constant which is based on the Charles law. Pressure is proportional to the temperature. Therefore, the pressure of the gas increases....
Read More →Find the equations of the tangent
Question: Find the equations of the tangent and the normal to the following curves at the indicated points. (i) $y=x^{4}-b x^{3}+13 x^{2}-10 x+5$ at $(0,5) \quad$ [NCERT] (ii) $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at $x=1 \quad$ [NCERT, CBSE 2011] $\begin{array}{ll}\text { (iii) } y=x^{2} \text { at }(0,0) \text { [NCERT] }\end{array}$ (iv) $y=2 x^{2}-3 x-1$ at $(1,-2)$ (v) $y^{2}=\frac{x^{3}}{4-x}$ at $(2,-2)$ (vi) $y=x^{2}+4 x+1$ at $x=3 \quad$ [CBSE 2004] (vii) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b...
Read More →Prove that
Question: Prove that $\frac{\sin x}{1+\cos x}=\tan \frac{x}{2}$ Solution: To Prove: $\frac{\sin x}{1+\cos x}=\tan \frac{x}{2}$ Proof: consider, L.H.S $=\frac{\sin x}{1+\cos x}$ $\frac{\sin x}{1+\cos x}=\frac{2 \cos \frac{x}{2} \sin \frac{x}{2}}{1+\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}\left(\because \cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}=\cos x\right.$ and $\left.2 \cos \frac{x}{2} \sin \frac{x}{2}=\sin x\right)$ $=\frac{2 \cos \frac{x}{2} \sin \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^...
Read More →Prove that
Question: Prove that $\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=2 \sec x$ Solution: To prove: $\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=2 \sec x$ Proof: Consider, L.H.S $=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)$ $\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\p...
Read More →Prove that
Question: Prove that $\sqrt{\frac{1+\sin x}{1-\sin x}}=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$ Solution: To Prove: $\sqrt{\frac{1+\sin x}{1-\sin x}}=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$ Proof: Consider, L.H.S $=\sqrt{\frac{1+\sin x}{1-\sin x}}$ Multiply and divide L.H.S by $\sqrt{1+\sin x}$ $=\sqrt{\frac{1+\sin x}{1-\sin x}} \times \frac{\sqrt{1+\sin x}}{\sqrt{1+\sin x}}=\frac{1+\sin x}{\sqrt{1-\sin ^{2} x}}$ $=\frac{1+\sin x}{\cos x}=\frac{1+2 \cos \frac{x}{2} \sin \frac{x}{2}}{\co...
Read More →Prove that
Question: Prove that $\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\tan x+\sec x$ Solution: To Prove: $\tan \left(\frac{\pi}{4}+\frac{\pi}{2}\right)=\tan x+\sec x$ Proof: Consider L.H.S, $\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{x}{2}}\left(\because\right.$ this is of the form $\left.\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right)$ $=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}=\frac{1+\frac{\sin \frac{x}{2}}{\...
Read More →Prove that
Question: Prove that $\cot \frac{x}{2}-\tan \frac{x}{2}=2 \cot x$ Solution: To Prove: $\cot \frac{x}{2}-\tan \frac{x}{2}=2 \cot x$ Proof: Consider L.H.S, $\cot \frac{x}{2}-\tan \frac{x}{2}=\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}-\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$ $=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\sin \frac{x}{2} \cos \frac{x}{2}}$ $=\frac{\cos x}{\sin \frac{x}{2} \cos \frac{x}{2}}\left(\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right)$ $\Rightarrow\left(\cos ^{2} \fr...
Read More →Find the equation of the normal
Question: Find the equation of the normal to $y=2 x^{3}-x^{2}+3$ at $(1,4)$. Solution: $y=2 x^{3}-x^{2}+3$ Differentiating both sides w.r.t. $x$, $\frac{d y}{d x}=6 x^{2}-2 x$ Slope of tangent $=\left(\frac{d y}{d x}\right)_{(1,4)}=6(1)^{2}-2(1)=4$ Slope of normal $=\frac{-1}{\text { Slope of tangent }}=\frac{-1}{4}$ Given $\left(x_{1}, y_{1}\right)=(1,4)$ Equation of normal is, $y-y_{1}=m\left(x-x_{1}\right)$ $\Rightarrow y-4=\frac{-1}{4}(x-1)$ $\Rightarrow 4 y-16=-x+1$ $\Rightarrow x+4 y=17$...
Read More →Solve this
Question: If $\sin x=\frac{3}{5}$ and $0x\frac{\pi}{2}$, find the value of $\tan \frac{x}{2}$ Solution: Given: $\sin x=\frac{3}{5}$ and $0x\frac{\pi}{2}$ i.e, $x$ lies in Quadrant $I$ and all the trigonometric ratios are positive in quadrant I To Find: $\tan \frac{X}{2}$ Formula used: $\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$ Now, $\cos x=\sqrt{1-\sin ^{2} x}(\because \cos x$ is positive in I quadrant) $\Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$ ...
Read More →Find the equation of the tangent to the curve
Question: Find the equation of the tangent to the curve $\sqrt{x}+\sqrt{y}=a$, at the point $\left(\frac{a^{2}}{4}, \frac{a^{2}}{4}\right)$. Solution: $\sqrt{x}+\sqrt{y}=a$ Differentiating both sides w.r.t. $x$, $\Rightarrow \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$ $\Rightarrow \frac{d y}{d x}=\frac{-\sqrt{y}}{\sqrt{x}}$ Given $\left(x_{1}, y_{1}\right)=\left(\frac{a^{2}}{4}, \frac{a^{2}}{4}\right)$ Slope of tangent, $m=\left(\frac{d y}{d x}\right)\left(\frac{a^{2}}{4}, \frac...
Read More →Is it possible to increase the temperature
Question: Is it possible to increase the temperature of a gas without adding heat to it? Explain. Solution: Yes, it is possible to increase the temperature of a gas without adding heat to it. This is possible during adiabatic compression....
Read More →If a refrigerator’s door is kept open,
Question: If a refrigerators door is kept open, will the room becomes cool or hot? Explain. Solution: If a refrigerators door is kept open, the room becomes hot because the amount of heat absorbed by the refrigerator and the work done by the refrigerator will be rejected to the room....
Read More →Solve this
Question: If $\operatorname{Cos} \frac{X}{2}=\frac{12}{13}$ and $X$ lies in Quadrant $I$, find the values of (i) $\sin x$ (ii) $\cos x$ (iii) $\cot x$ Solution: Given: $\cos \frac{X}{2}=\frac{12}{13}$ and $\mathrm{x}$ lies in Quadrant $\mathrm{I}$ i.e, All the trigonometric ratios are positive in I quadrant To Find: (i) $\sin x$ ii) $\cos x$ iii) $\cot x$ (i) $\sin x$ Formula used: We have, $\operatorname{Sin} x=\sqrt{1-\cos ^{2} x}$ We know that, $\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}(\bec...
Read More →Find the points on the curve
Question: Find the points on the curve $y=x^{3}$ where the slope of the tangent is equal to the $x$-coordinate of the point. Solution: Let $\left(x_{1}, y_{1}\right)$ be the required point. $x$ coordinate of the point is $x_{1}$. Since, the point lies on the curve. Hence, $y_{1}=x_{1}{ }^{3}$ ....(1) Now, $y=x^{3}$ $\Rightarrow \frac{d y}{d x}=3 x^{2}$ Slope of tangent at $(x, y)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 x_{1}^{2}$ Given that Slope of tangent at $\left(x_{1}, y_...
Read More →Can a system be heated and
Question: Can a system be heated and its temperature remains constant? Solution: For temperature to remain constant, the work done by the system against the surrounding should compensate with the heat that is supplied. According to the given information, $\Delta T=0 \Rightarrow \Delta U=0$ Therefore, $\Delta Q=\Delta U+\Delta W \Rightarrow \Delta Q=\Delta W$...
Read More →An ideal gas undergoes isothermal process
Question: An ideal gas undergoes isothermal process from some initial state I to final state f. Choose the correct alternatives (a) dU = 0 (b) dQ = 0 (c) dQ = dU (d) dQ = dW Solution: The correct answers are (a) dU = 0 (d) dQ = dW...
Read More →Which of the process described below are irreversible?
Question: Which of the process described below are irreversible? (a) the increase in temperature of an iron rod by hammering it (b) a gas in a small container at a temperature T1is brought in contact with a big reservoir at a higher temperature T2which increases the temperature of the gas (c) a quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless piston (d) an ideal gas is enclosed in a piston-cylinder arrangement with adiabatic walls. A weight W is added to...
Read More →Who that the tangents to the curve
Question: Who that the tangents to the curve $y=7 x^{3}+11$ at the points $x=2$ and $x=-2$ are parallel. Solution: Given: $y=7 x^{3}+11$ $\therefore \frac{d y}{d x}=21 x^{2}$ Now, Slope of the tangent at $(x=2)=\left(\frac{d y}{d x}\right)_{x=2}=21(2)^{2}=84$ Slope of the tangent at $(x=-2)=\left(\frac{d y}{d x}\right)_{x=-2}=21(-2)^{2}=84$ Both slopes are the same. Hence, the tangents at points $x=2$ and $x=-2$ are parallel....
Read More →Three copper blocks of masses M1,
Question: Three copper blocks of masses M1, M2, and M3kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T1, T2, T3(T1 T2 T3). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (a) $T=\frac{T_{1}+T_{2}+T_{3}}{3}$ (b) $T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{M_{1}+M_{2}+M_{3}}$ (c) $T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{3\left(M_{1}+M_{2}+M_{3}\right)}$ (d) $T=\frac{M_{1} T_{1} s+M_{2} T_{2} s+M...
Read More →Consider two containers A and B containing
Question: Consider two containers A and B containing identical gases at the same pressure, volume, and temperature. The gas in container A is compressed to half of its volume isothermally while the gas in the container B is compressed to half of its original value adiabatically. The ratio of final pressure of a gas in B to that of gas in A is (a) $2^{\gamma-1}$ (b) $\left(\frac{1}{2}\right)^{\gamma-1}$ (c) $\left(\frac{1}{1-\gamma}\right)^{2}$ (d) $\left(\frac{1}{\gamma-1}\right)^{2}$ Solution: ...
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