If A + B + C = π, prove that
Question: If A + B + C = , prove that $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1-2 \cos A \cos B \cos C$ Solution: $=\cos ^{2} A+\cos ^{2} B+\cos ^{2} C$ Using formula, $\frac{1+\cos 2 A}{2}=\cos ^{2} A$ $=\frac{1+\cos 2 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}+\frac{1+\cos 2 \mathrm{C}}{2}$ $=\frac{1+\cos 2 \mathrm{~A}+1+\cos 2 \mathrm{~B}+1+\cos 2 \mathrm{C}}{2}$ $=\frac{3+\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}}{2}$ Using, $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \co...
Read More →Find the values of a and b
Question: Find the values of $a$ and $b$ if the The Slope of the tangent to the curve $x y+a x+b y=2$ at $(1,1)$ is 2 . Solution: Given: The Slope of the tangent to the curve $x y+a x+b y=2$ at $(1,1)$ is 2 First, we will find The Slope of tangent we use product rule here, $\therefore \frac{\mathrm{d}}{\mathrm{dx}}(U V)=U \times \frac{\mathrm{dV}}{\mathrm{dx}}+\mathrm{V} \times \frac{\mathrm{dU}}{\mathrm{dx}}$ $\Rightarrow x y+a x+b y=2$ $\Rightarrow x \times \frac{d}{d x}(y)+y \times \frac{d}{d...
Read More →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 : $x y=6$ at $(1,6)$ Solution: Given: $x y=56$ at $(1,6)$ Here we have to use the product rule for above equation. If $u$ and $v$ are differentiable function, then $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{UV})=\mathrm{U} \times \frac{\mathrm{dV}}{\mathrm{dx}}+\mathrm{V} \times \frac{\mathrm{dU}}{\mathrm{dx}}$ $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\frac{\mathrm{d}}{\mathrm{dx}}(6)$ $\...
Read More →The displacement of a particle is represented
Question: The displacement of a particle is represented by the equation y = 3 cos(/4 2t). The motion of the particle is (a) simple harmonic with period 2p/w (b) simple harmonic with period / (c) periodic but not simple harmonic (d) non-periodic Solution: The correct answer is (b) simple harmonic with period /...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\frac{\cos A}{\sin B \sin C}+\frac{\cos B}{\sin C \sin A}+\frac{\cos C}{\sin A \sin B}=2$ Solution: $=\frac{\cos A}{\sin B \sin C}+\frac{\cos B}{\sin C \sin A}+\frac{\cos C}{\sin A \sin B}$ Taking L.C.M $=\frac{\cos A \sin A+\cos B \sin B+\cos C \sin C}{\sin B \sin C \sin A}$ Multiplying and divide the above equation by 2, we get $=\frac{2 \cos A \sin A+2 \cos B \sin B+2 \cos C \sin C}{2 \sin B \sin C \sin A}$ Since , sin2A = 2sinAcosA $=\frac{\sin 2 A+\sin...
Read More →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 : $x^{2}+3 y+y^{2}=5$ at $(1,1)$ Solution: Given: $x^{2}+3 y+y^{2}=5$ at $(1,1)$ Here we have to differentiate the above equation with respect to $x$. $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+3 \mathrm{y}+\mathrm{y}^{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(5)$ $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dx}}(3 \ma...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\sin (B+C-A)+\sin (C+A-B)-\sin (A+B-C)=4 \cos A \cos B \sin C$ Solution: $=\sin (B+C-A)+\sin (C+A-B)-\sin (A+B-C)$ Using, $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $=2 \sin C \cos (B-A)-\sin (A+B-C)$ Since $A+B+C=\pi$ $\rightarrow \mathrm{B}+\mathrm{A}=180-\mathrm{C}$ $=2 \sin C \cos (B-A)-\sin (\pi-C-C)$ $=2 \sin C \cos (B-A)-\sin 2 C$ Since , sin2A = 2sinAcosA, $=2 \sin C \cos (B-A)-2 \sin C \cos C$ $=2 \sin C\{\cos...
Read More →If the solve the problem
Question: $y=(\sin 2 x+\cot x+2)^{2}$ at $x=\pi / 2$ Solution: Given: $y=(\sin 2 x+\cot x+2)^{2} a t x=\frac{\pi}{2}$ First, we have to find $\frac{d y}{d x}$ of given function, $f(x)$,i.e, to find the derivative of $f(x)$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$ The Slope of the tangent is $\frac{d y}{d x}$ $\Rightarrow y=(\sin 2 x+\cot x+2)^{2}$ $\frac{d y}{d x}=2 x(\sin 2 x+\cot x+2)^{2}-1\left\{\frac{d y}{d x}...
Read More →Consider a rectangular block of wood moving
Question: Consider a rectangular block of wood moving with a velocity vo in a gas at temperature T and mass density . Assume the velocity is along the x-axis and the area of cross-section of the block perpendicular to vo is A. Show that the drag force on the block is $4 \rho A v_{0} \sqrt{\frac{k T}{m}}$, where $m$ is the mass of the gas molecule. Solution: Let m is the no.of molecules per unit volume Change in momentum by a molecule on front side = 2m (v + v0) Change in momentum by a molecule o...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ Solution: = sin2A + sin2B + sin2C Using, $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\sin 2 A=2 \sin A \cos A$ $=2 \sin A \cos A+2 \sin (B+C) \cos (B-C)$ since $A+B+C=\pi$ $\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$ $=2 \sin A \cos A+2 \sin (\pi-A) \cos (B-C)$ $=2 \sin A \cos A+2 \sin A \cos (B-C)$ $=2 \...
Read More →A box of 1.00 m3 is filled with nitrogen
Question: A box of 1.00 m3 is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area 0.010 mm2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm? Solution: Volume of the box = 1 m3= V1 Initial pressure P1 = 1.5 atm Final pressure P2 = 1.4 atm Air pressure Pa = 1 atm Initial temperature T1 = 300 K Final temperature T2 = 300 K Area of the hole = 10-8m2 Pressure difference between tyre and atmosphere = 1.5 1 atm Mass of nitrogen = ...
Read More →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 : $x=a(\theta-\sin \theta), y=a(1-\cos \theta)$ at $\theta=\pi / 2$ Solution: Here, To find $\frac{d y}{d x}$, we have to find $\frac{d y}{d \theta} \ \frac{d x}{d \theta}$ and and divide $\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$ and we get our desired $\frac{d y}{d x}$. $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\...
Read More →Ten small planners are flying at a speed of 150 km/h
Question: Ten small planners are flying at a speed of 150 km/h in total darkness in an air space that is 20 20 1.5 km3in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m. Solution: Time = distance/speed No.of particle...
Read More →Consider an ideal gas with
Question: Consider an ideal gas with the following distribution of speeds (a) calculate Vrms and hence T (m = 3.0 10-26kg) (b) if all the molecules with speed 1000 m/s escape from the system, calculate new Vrmsand hence T Solution: (a) T = 296 K (b) T = 248.04 K...
Read More →Explain why
Question: Explain why (a) there is no atmosphere on moon (b) there is a fall in temperature with altitude Solution: (a) There is no atmosphere on the moon because the gravitational force is small and the Vrms is the greater on the moon such that the escape velocity of the molecule is greater than that of the air. Also, the distance between the moon and the sun is the same as the distance between the moon and the earth. Therefore, the rms speed of the molecule increases such that it can be more t...
Read More →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 $f(x)$ $y=\sqrt{x^{3}}$ $\therefore \sqrt[n]{x}=x^{\frac{1}{n}}$ $\Rightarrow y=\left(x^{3}\right)^{\frac{1}{2}}$ $\Rightarrow y=(x)^{\frac{3}{2}}$ $\therefore \frac{d y}{d x}\left(x^{n}\right)=n \cdot x^{n-1}$ The Slope of...
Read More →An insulated container containing monoatomic
Question: An insulated container containing monoatomic gas of molar mass m is moving with a velocity vo. If the container is suddenly stopped, find the change in temperature. Solution: The final KE of the gas = 0 Change in KE, ∆K= 1/2 (nm)v2 ∆T is the change in the temperature ∆U = nCv∆T ∆K = ∆U ∆T = mv02/3R...
Read More →Calculate the number of degrees
Question: Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP. Solution: The volume occupied by the molecules of gas = 22400 cc No.of molecules in 1 cc of hydrogen = 2.688 1019 Hydrogen has a total of 5 degrees of free as it is a diatomic gas Therefore, the total degrees of freedom = 1.344 1020...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\cos A+\cos B+\cos C=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ Solution: = cosA + cosB + cosC Using $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $=\cos A+\left\{2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}$ since A + B + C = $\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$ And, $\cos \left(\frac{\pi}{2}-A\right)=\sin A$ $=\cos A+\left\{2 \cos \left(\frac{\pi-A}{2}\right) \cos ...
Read More →A balloon has 5.0 g mole of
Question: A balloon has 5.0 g mole of helium at 7oC. Calculate (a) the number of atoms of helium in the balloon (b) the total internal energy of the system Solution: Average KE per molecule = 3/2kT No.of moles of helium, n = 5 g mole T = 280 K a) No.of atoms of helium in the balloon = 30.015 1023 b) KE = 3/2 kbT Total internal energy = 1.7 104J...
Read More →A balloon has 5.0 g mole of
Question: A balloon has 5.0 g mole of helium at 7oC. Calculate (a) the number of atoms of helium in the balloon (b) the total internal energy of the system Solution: Average KE per molecule = 3/2kT No.of moles of helium, n = 5 g mole T = 280 K (a) No.of atoms of helium in the balloon = 30.015 1023 (b) KE = 3/2 kbT Total internal energy = 1.7 104J...
Read More →When air is pumped into a cycle tyre
Question: When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyles law in this case? Solution: PV = P(m/) = constant P/ = constant Volume = m/ where m is constant Therefore, when the air is pumped into the tyre of the cycle, the mass of air increases as the no.of molecules increases. Therefore, Boyles law is only applicable when the mass of the gas remains fixed....
Read More →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: $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\pi / 4$ Solution: Given: $x=\operatorname{acos}^{3} \theta \ y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$ Here, To find $\frac{d y}{d x}$, we have to find $\frac{d y}{d \theta} \ \frac{d x}{d \theta}$ and and divide $\frac{\frac{d y}{d \theta}}{d \theta}$ and we get our desired $\frac{d y}{d x}$. $\therefore \frac{\mathrm{dy}}{\m...
Read More →We have 0.5 g of hydrogen gas in a cubic
Question: We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? Solution: Volume of 1 molecule = 4/3 r3 = 4.20 10-30m3 No.of moles in 0.5 g H2 gas = 0.25 mole Volume of H2 molecule in 0.25 mole = 1.046.023 10+23-30...
Read More →If A + B + C = π, prove that
Question: If A + B + C = , prove that $\sin A+\sin B+\sin C=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ Solution: = sinA + sinB + sinC Using, $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $=\sin A+\left\{2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}$ Since $A+B+C=\pi$ $\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$ And, $\sin \left(\frac{\pi}{2}-\mathrm{A}\right)=\cos \mathrm{A}$ $=\sin A+\left\{2 \sin \left(\frac{\pi-A...
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