Which of the following sentences are statements?
Question: Which of the following sentences are statements? In case of a statement mention whether it is true or false. (i) The sun is a star. (ii) $\sqrt{7}$ is an irrational number. (iii) The sum of 5 and 6 is less than 10 . (iv) Go to your class. (v) Ice is always cold. (vi) Have you ever seen the Red Fort? (vii) Every relation is a function. (viii) The sum of any two sides of a triangle is always greater than the third side. (ix) May God bless you! Solution: (i) The sun is a star is a stateme...
Read More →Calculate potential on the axis of a disc
Question: Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface. Solution: In the above figure, we can see that the disc is divided into several charged rings. Let P be the point on the axis of the disc at a distance x from the centre of the disc. The radius of the ring is r and the width is dr. dq is the charge on the ring which is given as dq = dA = 2rdr The potential is given as $d V=\frac{1}{4 \pi \epsilon_{0}} \frac{d q}{\sqrt{r^{2}+x^{...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x$ Solution: let $I=\int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x$ Let $\tan x=t \ldots \ldots(i)$ $\Rightarrow \sec ^{2} x d x=d t$ SO, $I=\int \frac{d t}{(1)^{2}-t^{2}}$ $\mathrm{I}=\frac{1}{2 \times 1} \log \left|\frac{1+\mathrm{t}}{1-\mathrm{t}}\right|+\mathrm{c}\left[\right.$ since, $\left.\int \frac{1}{\mathrm{a}^{2}-(\mathrm{x})^{2}} \mathrm{dx}=\frac{1}{2 \times \mathrm{a}} \log \left|\frac{\mathrm{a}+\mat...
Read More →In the circuit shown in the figure, initially, K1 is closed and K2
Question: In the circuit shown in the figure, initially, K1is closed and K2is open. What are the charges on each capacitor. Then K1was opened and K2was closed. What will be the charge on each capacitor now? Solution: When K1is closed and K2is open, the capacitors C1and C2are connected in series with the battery Therefore, the charge in capacitors C1and C2are Q1= Q2= q = (C1C2/C1+C2)E = 18C When the capacitors C2and C3are placed in parallel, C2V + C3V = Q2 V = Q2/C2+ C3= 3V Therefore, Q2 = 3CV = ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{x^{2}+6 x+13} d x$ Solution: We have, $x^{2}+6 x+13=x^{2}+6 x+3^{2}-3^{2}+13$ $=(x+3)^{2}+4$ Sol, $\int \frac{1}{x^{2}+6 x+13} d x=\int \frac{1}{(x+3)^{2}+2^{2}} d x$ Let $x+3=t$ Then $d x=d t$ $\int \frac{1}{(\mathrm{t})^{2}+2^{2}} \mathrm{dt}=\frac{1}{2} \tan ^{-1} \frac{\mathrm{t}}{2}+\mathrm{c}$ [since, $\left.\int \frac{1}{x^{2}+(a)^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right]$ $\frac{1}{2} \tan ^{-1} \frac{x+3}{...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{2 x^{2}-x-1} d x$ Solution: let $I=\int \frac{1}{2 x^{2}-x-1} d x$ $=\frac{1}{2} \int \frac{1}{x^{2}-\frac{x}{2}-\frac{1}{2}} d x$ $=\frac{1}{2} \int \frac{1}{x^{2}+2 x \times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}-\frac{1}{2}} d x$ $=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}} d x$ Let $\left(x-\frac{1}{4}\right)=t \ldots \ldots$ (i) $\Rightarrow \mathrm{d} x=\mathrm{dt}$ SO, $I=...
Read More →Solve this
Question: $y=\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}$, show that $\frac{d y}{d x}=\sec x(\tan x+\sec x)$ Solution: $y=\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}$ $y=\sqrt{\frac{\frac{1}{\cos x} \frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}}=\sqrt{\frac{1-\sin x}{1+\sin x}}$ $u=1-\sin x, v=1+\sin x, z=\frac{1-\sin x}{1+\sin x}$ Formula : $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ According to quotient rule of differentiation If $z=\frac{u}{v}$ $\mathrm{d...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{1+x-x^{2}} d x$ Solution: : let $\mathrm{I}=\int \frac{1}{1+\mathrm{x}-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\int \frac{1}{-\left(\mathrm{x}^{2}-\mathrm{x}-1\right)} \mathrm{dx}$ $=\int \frac{1}{-\left(x^{2}-x-1\right)} d x$ $=\int \frac{1}{-\left(x^{2}-x-\frac{1}{4}-1+\frac{1}{4}\right)} d x$ $=\int \frac{1}{-\left(\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}\right)} d x$ $=\int \frac{1}{\left(\left(\frac{\sqrt{5}}{2}\right)^{2}-\left(x-...
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Question: $y=\sqrt{\frac{1-x}{1+x}}$, prove that $\left(1-x^{2}\right) \frac{d y}{d x}+y=0$ Solution: Let $y=\sqrt{\frac{1-x^{1}}{1+x^{1}}}, u=1-x^{1}, v=1+x^{1}, z=\frac{1-x^{1}}{1+x^{1}}$ According to quotient rule of differentiation If $\mathrm{z}=\frac{\mathrm{u}}{\mathrm{V}}$ $\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$ $=\frac{\left(1+x^{1}\right) \times(-1)-\left(1-x^{1}\right) \times...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{x^{2}-10 x+34} d x$ Solution: let $I=\int \frac{1}{x^{2}-10 x+34} d x$ $I=\int \frac{1}{x^{2}-10 x+34} d x$ $=\int \frac{1}{x^{2}+2 x \times 5+(5)^{2}-(5)^{2}+34} d x$ $=\int \frac{1}{(x-5)^{2}-9} d x$ Let $(x-5)=t \ldots \ldots$ (i) $\Rightarrow d x=d t$ So, $I=\int \frac{1}{t^{2}+(3)^{2}} d t$ $I=\frac{1}{3} \tan ^{-1}\left(\frac{t}{3}\right)+c$ $\left[\right.$ since, $\left.\int \frac{1}{x^{2}+(a)^{2}} d x=\frac{1}{a} \tan ^{-1}\left(...
Read More →show that
Question: If $y=\frac{\cos x+\sin x}{\cos x-\sin x}$, show that $\frac{d y}{d x}=\sec ^{2}\left(x+\frac{\pi}{4}\right)$ Solution: Let $y=\frac{\cos x+\sin x}{\cos x-\sin x}, u=\cos x+\sin x, v=\cos x-\sin x$ Formula: $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ and $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$ According to the quotient rule of differentiation If $y=\frac{u}{v}$ $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{4 x^{2}+12 x+5} d x$ Solution: let $I=\int \frac{1}{4 x^{2}+12 x+5} d x$ $=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{4}} d x$ $=\frac{1}{4} \int \frac{1}{x^{2}+2 x \times \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}} d x$ $=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-1} d x$ Let $\left(\mathrm{x}+\frac{3}{2}\right)=\mathrm{t} \ldots \ldots$ (i) $\Rightarrow \mathrm{d} \mathrm{x}=\mathr...
Read More →Find the value
Question: If $y=\frac{\cos x-\sin x}{\cos x+\sin x}$, show that $\frac{d y}{d x}+y^{2}+1=0$ Solution: $=-\frac{1}{1}-y^{2}\left(y=\frac{\cos x-\sin x}{\cos x+\sin x}\right)$ Formula: $\frac{d(\sin x)}{d x}=\cos x$ and $\frac{d(\cos x)}{d x}=-\sin x$ According to the quotient rule of differentiation If $y=u / v$ $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$ $=\frac{(\cos x+\sin x) \times(-\sin...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{4}+1}{x^{2}+1} d x$ Solution: We will use basic formula : $(a+b)^{2}=a^{2}+b^{2}+2 a b$ Or, $a^{2}+b^{2}=(a+b)^{2}-2 a b$ Here, $x^{4}+1=x^{4}+1^{4}$ $=\left(\mathrm{x}^{2}\right)+\left(1^{2}\right)^{2}$ Applying above formula, we get, $x^{4}+1=\left(x^{2}+1\right)^{2}-2 \times 1 \times x^{2}$ $=\left(x^{2}+1\right)^{2}-2 x^{2}$ Hence, $\int \frac{x^{4}+1}{x^{2}+1} d x=\int \frac{\left(x^{2}+1\right)^{2}-2 x^{2}}{x^{2}+1} d x$ Separate t...
Read More →Find the value
Question: Find $\frac{d y}{d x}$, When $y=\frac{\sin x+x^{2}}{\cot 2 x}$ Solution: Let $y=\frac{\sin x+x^{2}}{\cot 2 x}, u=\sin x+x^{2}, v=\cot 2 x$ Formula: $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}, \frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{n} \times \mathrm{x}^{\mathrm{n}-1}$ and $\frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec}^{2} \mathrm{x}$ According to the quotient rule of differentiation If $y=\frac{u}{v}$ $\ma...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{(2-x)^{2}-1}} d x$ Solution: Let $(2-x)=t$, then $d t=-d x$, or $d x=-d t$ Hence, $\int \frac{1}{\sqrt{(2-x)^{2}-1}} d x=\int \frac{1}{t^{2}-1}(-d t)$ $\left.=-\int \frac{1}{t^{2}-1^{2}} d t=-\log \int\left(t+\sqrt{t^{2}-1}\right)\right)+c\left\{\right.$ since $\left.\int \frac{1}{\left.\sqrt{(} x^{2}+a^{2}\right)} d x=\log \left[x+\sqrt{\left(x^{2}-a^{2}\right)}+c\right\}\right\}$ Put $t=2-x$ $\left.=-\log \int\left((2-x)+\sqrt{(2...
Read More →Two metal spheres, one of radius R and the other of radius 2R,
Question: Two metal spheres, one of radius R and the other of radius 2R, both have the same surface charge density . They are brought in contact and separated. What will be new surface charge densities on them? Solution: Following are the charges on the metal sphere before contact Q1= .4R2 Q2= .4(2R)2= 4Q1 Q1 and Q2 are the charges on the metal sphere after contact Q1 + Q2 = Q1+ Q2= 5Q1 When the metal spheres are in contact, the following is the potentials acquired by them Q1 = Q2/2 Solving the ...
Read More →Find the value
Question: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, When $\quad \mathrm{y}=\sin \left(\frac{1+\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right)$ Solution: Let $y=\sin \left(\frac{1+x^{2}}{1-x^{2}}\right), u=1+x^{2}, v=1-x^{2}, z=\frac{1+x^{2}}{1-x^{2}}$ Formula : $\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=2 \mathrm{x}$ and $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ According to the quotient rule of differentiation If $z=\frac{u}{v}$ $\mathrm{dz} / \mathrm{dx}=\frac{\m...
Read More →(a) In a quark model of elementary particles,
Question: (a) In a quark model of elementary particles, a neutron is made of one up quarks and two down quarks. Assume that they have a triangle configuration with side length of the order of 10-15m. Calculate the electrostatic potential energy of neutron and compare it with its mass 939 MeV. (b) Repeat above exercise for a proton which is made of two up and one down quark. Solution: There are three charges in the system. The potential energy of the system is equal to the sum of the PE of each p...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{(2-x)^{2}+1}} d x$ Solution: Let $(2-x)=t$, then $d t=-d x$, or $d x=-d t$ Hence, $\int \frac{1}{\sqrt{(2-x)^{2}+1}} d x=\int \frac{1}{t^{2}+1}(-d t)$ $\left.=-\int \frac{1}{t^{2}+1^{2}} d t=-\log \int\left(t+\sqrt{t^{2}+1}\right)\right)+c\left\{\right.$ since $\left.\left.\int \frac{1}{\sqrt{\left(a^{2}+x^{2}\right)}} d x=\log \left[x+\sqrt{\left(a^{2}\right.}+x^{2}\right)+c\right\}\right\}$ Put $t=2-x$ $\left.=-\log \int\left((2-...
Read More →A capacitor is made of two circular plates
Question: A capacitor is made of two circular plates of radius R each, separated by a distance d R. The capacitor is connected to a constant voltage. A thin conducting disc of radius r R and thickness t r is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m. Solution: When the conducting disc is placed at the centre of the bottom plate, the potential of the disc is equal to the potential of the plate. The electric field on the...
Read More →A parallel plate capacitor is filled by a dielectric
Question: A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as = U where = 2V-1. A similar capacitor with no dielectric is charged to U0= 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. Solution: Since the capacitors are connected in parallel, the potential difference across the capacitors is the same. The final voltage is assumed to be U. C is the capacitance o...
Read More →Prove that, if an insulated, uncharged conductor
Question: Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity. Solution: The electric potential decreasing along the direction of the electric field is given as E = dV/dr The electric potential decreases when the path from the charged conductor is taken to the uncharged conductor along the direction of the electric field. ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{a^{2}-b^{2} x^{2}}} d x$ Solution: Let $\mathrm{bx}=\mathrm{t}$ then $\mathrm{dt}=\mathrm{bdx}$ or $\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{b}}$ Hence, $\int \frac{1}{\sqrt{a^{2}-b^{2} x^{2}}} d x=\frac{1}{b} \int \frac{1}{\sqrt{\left(a^{2}-t^{2}\right)}} d t$ $=\frac{1}{b} \int \sin ^{-1}\left(\frac{t}{a}\right)+c\left\{\right.$ since $\left.\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right\}$ Put $t=...
Read More →Prove that a closed equipotential surface
Question: Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume. Solution: In a closed equipotential surface, the potential changes from position to position. The potential inside the surface is different from the potential gradient caused in the surface that is dV/dr This also means that electric field is not equal to zero and it is given as E = -dV/dr Therefore, it could be said that the field lines are either pointing inwards or outwards t...
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