A parallel plate capacitor is made of two dielectric blocks in series.
Question: A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d1and dielectric constant k1and the other has thickness d2and dielectric constant k2as shown in the figure. This arrangement can be thought of as a dielectric slab of thickness d = d1+ d2and effective dielectric constant k. The k is (a) k1d1+ k2d2/d1+d2 (b) k1d1+ k2d2/k1+ k2 (c) k1k2(d1+ d2)/(k1d1+ k2d2) (d) 2k1k2/k1+ k2 Solution: The correct answer is (c) k1k2(d1+ d2)/(k1d1+ k2d2)...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t x: $\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$ Solution: Let $y=\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}, u=a^{2}-x^{2}, v=a^{2}+x^{2}, z=\frac{a^{2}-x^{2}}{a^{2}+x^{2}}$ Formula : $\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}=2 \mathrm{x}$ According to the 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{...
Read More →Equipotential at a great distance from a collection
Question: Equipotential at a great distance from a collection of charges whose total sum is not zero are approximately (a) spheres (b) planes (c) paraboloids (d) ellipsoids Solution: The correct answer is (a) spheres...
Read More →The electrostatic potential on the surface of a charged conducting
Question: The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statements are made in this regard: S1: At any point inside the sphere, the electric intensity is zero S2: At any point inside the sphere, the electrostatic potential is 100V Which of the following is a correct statement? (a) S1is true but S2is false (b) Both S1and S2are false (c) S1is true, S2is also true, and S1is the cause of S2 (d) S1is true, S2is also true but the statements are independent Solu...
Read More →Figure shows some equipotential lines distributed
Question: Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B. (a) the work done in fig (i) is the greatest (b) the work done in fig (ii) is least (c) the work done is the same in fig (i), fig (ii), and fig (iii) (d) the work done in fig (iii) is greater than fig (ii) but equal to that in fig (i) Solution: The correct answer is (c) the work done is the same in fig (i), fig (ii), and fig (iii)...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t x: $\sqrt{\frac{1-x^{2}}{1+x^{2}}}$ Solution: Let $y=\sqrt{\frac{1-x^{2}}{1+x^{2}}}, u=1-x^{2}, v=1+x^{2}, z=\frac{1-x^{2}}{1+x^{2}}$ Formula : $\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=2 \mathrm{x}$ According to the quotient rule of differentiation If $\mathrm{z}=\frac{\mathrm{u}}{\mathrm{v}}$ $\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$ $=\fra...
Read More →A positively charged particle is released
Question: A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge (a) remains a constant because the electric field is uniform (b) increases because the charge moves along the electric field (c) decreases because the charge moves along the electric field (d) decreases because the charge moves opposite to the electric field Solution: The correct answer is (c) decreases because the charge moves along the electric field...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t x: $\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}$ Solution: Let $y=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}, u=e^{2 x}+e^{-2 x}, v=e^{2 x}-e^{-2 x}$ Formula : $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$ According to the quotient rule of differentiation If $y=\frac{u}{v}$ $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$ ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x$ Solution: let $x=\tan \theta$, so $d x=\sec ^{2} \theta d \theta$ and $\theta=\tan ^{-} x$ Putting above values, $=\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x=\int \frac{\sqrt{1+\tan ^{2} \theta}}{\tan ^{4} \theta} \sec ^{2} \theta d \theta=\int \sec ^{2} \theta / \tan ^{2} \theta d \theta$ $=\int \operatorname{cosec}^{2} \theta d \theta=-\cot \theta+c$ Put $\theta=\tan ^{-} \mathrm{x}$ $=-\cot \theta+c=-\cot \tan ^{-}...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right\} d x$ Solution: Let $x=\cos 2 t$ and $t=\cos ^{-} x \frac{x}{2}$ $=\sqrt{\frac{1+x}{1-x}}=\sqrt{\frac{1+\cos 2 t}{1-\cos 2 t}}$ We know $1+\cos 2 t=2 \cos ^{2} t$ and $1-2 \cos 2 t=2 \sin ^{2} t$ Hence, $\sqrt{\frac{1+\cos 2 t}{1-\cos 2 t}}=\sqrt{\frac{\cos ^{2} t}{\sin ^{2} t}=\sqrt{\cot ^{2} t}}=\cot t$ Therefore, $\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right\} \mathrm{dx}=\in...
Read More →Differentiate w.r.t x
Question: Differentiate w.r.t x $\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$ Solution: Let $y=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}, u=e^{x}+e^{-x}, v=e^{x}-e^{-x}$ Formula : $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$ According to the quotient rule of differentiation If $\mathrm{y}=\frac{u}{v}$ $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$ $=\frac{\left(e^...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{7}}{\left(a^{2}-x^{2}\right)^{5}} d x$ Solution: PUT $x=a \sin \theta$, so $d x=a \cos \theta d \theta$ and $\theta=\sin ^{-}(x / a)$ Above equation becomes, $\int \frac{\mathrm{x}^{7}}{\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)^{5}} \mathrm{dx}==\int \frac{\mathrm{a}^{7} \sin ^{\prime} \theta}{\left(\mathrm{a}^{2}-\mathrm{a}^{2} \sin ^{2} \theta\right)^{5}}(\operatorname{acos} \theta \mathrm{d} \theta)=\int \frac{\mathrm{a}^{7} \sin ^{\...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t $x: e^{(x \sin x+\cos x)}$ Solution: Let $y=e^{(x \sin x+\cos x)}, z=x \sin x+\cos x, m=x$ and $w=\sin x$ Formula : $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}, \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ and $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$ According to the product rule of differentiation $\mathrm{dz} / \mathrm{dx}=\mathrm{w} \times \frac{\mathrm{dm}}{\mathrm{dx}}...
Read More →A capacitor of 4μF is connected as shown in the circuit.
Question: A capacitor of 4F is connected as shown in the circuit. The internal resistance of the battery is 0.5Ω. The amount of charge on the capacitor plates will be (a) 0 (b) 4C (c) 16C (d) 8C Solution: The correct answer is (d) 8C...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{3 / 2}} d x$ Solution: $\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{3 / 2}} d x$ PUT $x=a \sin \theta$, so $d x=a \cos \theta d \theta$ and $\theta=\sin ^{-}(x / a)$ Above equation becomes, $=\int \frac{a^{2} \sin ^{2} \theta}{\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{3 / 2}}(a \cos \theta d \theta)=\int \frac{a^{2} \sin ^{2} \theta}{\left(a^{2}\right)\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{3 / 2}}(a \co...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t $x: \cos \left(x^{3} \cdot e^{x}\right)$ Solution: Let $y=\cos \left(x^{3} \cdot e^{x}\right), z=x^{3} \cdot e^{x}, m=e^{x}$ and $w=x^{3}$ Formula : $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}, \frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{n} \times \mathrm{x}^{\mathrm{n}-1}$ and $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$ According to the product rule of differentiat...
Read More →Differentiate w.r.t x
Question: Differentiate w.r.t $x: e^{-5 x} \cot 4 x$ Solution: Let $y=e^{-5 x} \cot 4 x, z=e^{-5 x}$ and $w=\cot 4 x$ Formula $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$ and $\frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec}^{2} \mathrm{x}$ According to the product rule of differentiation $\mathrm{dy} / \mathrm{dx}=\mathrm{w} \times \frac{\mathrm{dz}}{\mathrm{dx}}+\mathrm{z} \times \frac{\mathrm{dw}}{\mathrm{dx}}$ $=\left[\cot ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sin x \cos ^{3} x} d x$We know, $\sin ^{2} x+\cos ^{2} x=1$ Solution: We know, $\sin ^{2} x+\cos ^{2} x=1$ Therefore $\frac{1}{\sin x \cos ^{2} x}=\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos ^{2} x}$ Divide each term of numerator separately by $\sin x \cos ^{3} x$ $=\frac{\sin ^{2} x}{\sin x \cos ^{3} x}+\frac{\cos ^{2} x}{\sin x \cos ^{2} x}=\frac{\sin x}{\cos ^{2} x}+\frac{1}{\sin x \cos x}$ $=\frac{\sin x}{\cos x} *\left(\frac{1}{\cos...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sin ^{3} x \cos x} d x$ Solution: $\int \frac{1}{\sin ^{3} x \cos x} d x=\int \sin ^{-3} x \cos ^{-1} x d x$ Adding the powers, $-3+-1=-4$ Since it is an even number, we will divide numerator and denominator by cosx $\int \frac{1}{\sin ^{3} x \cos x} d x=\int \frac{\frac{1}{\cos ^{2} x}}{\frac{\sin ^{2} x \cos x}{\cos ^{4} x}} d x$ $=\int \frac{\sec ^{4} x}{\tan ^{3} x} d x=\int \frac{\sec ^{2} x \sec ^{2} x}{\tan ^{3} x} d x$ $=\int \f...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sin ^{3} x \cos ^{5} x} d x$ Solution: $\int \frac{1}{\sin ^{3} x \cos ^{5} x} d x=\int \sin ^{-3} x \cos ^{-5} x d x$ Adding the powers, $-3+-5=-8$ Since it is an even number, we will divide numerator and denominator by $\cos ^{8} x$ $\int \frac{1}{\sin ^{3} x \cos ^{5} x} d x=\int \frac{\frac{1}{\cos ^{8} x}}{\frac{\sin ^{2} x \cos ^{5} x}{\cos ^{8} x}} d x$ $=\int \frac{\sec ^{8} x}{\tan ^{3} x} d x=\int \frac{\sec ^{6} x \sec ^{2} x...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: Solution: $\int \frac{1}{\sin ^{4} x \cos ^{2} x} d x$ Adding the powers : $-4+-2=-6$ Since all are even nos, we will divide each by $\cos ^{6} x$ to convert into positive power So, $\int \frac{1}{\sin ^{4} x \cos ^{2} x} d x=\int \frac{\frac{1}{\cos ^{6} x}}{\frac{\sin ^{4} x \cos ^{2} x}{\cos ^{6} x}} d x$ $=\int \frac{\sec ^{6} x}{\frac{\sin ^{4} x}{\operatorname{coc}^{4} x}} d x=\int \frac{\sec ^{6} x}{\tan ^{4} x} d x$ $=\int \frac{\sec ^{4} x \se...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \sin ^{3} x \cos ^{5} x d x$ Solution: Let $\cos x=t$ then $d t=-\sin x d x$ $\mathrm{dx}=-\frac{\mathrm{dt}}{\sin \mathrm{x}}$ Substitute all these in the above equation, $\int \sin ^{3} x \cos ^{5} x d x=\int \sin ^{3} x t^{5}\left(-\frac{d t}{\sin x}\right)$ $=-\int \sin ^{2} \mathrm{xt}^{5} \mathrm{dt}$ $=-\int\left(1-\cos ^{2} \mathrm{x}\right) \mathrm{t}^{5} \mathrm{dt}$ $=-\int\left(1-\mathrm{t}^{2}\right) \mathrm{t}^{5} \mathrm{dt}$ $=-\i...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \sin ^{7} x d x$ Solution: $\int \sin ^{7} x d x=\int \sin ^{6} x \sin x d x$ $=\int\left(\sin ^{2} x\right)^{3} \sin x d x$ $=\int\left(1-\cos ^{2} x\right)^{3} \sin x d x\left\{\right.$ since $\left.\sin ^{2} x+\cos ^{2} x=1\right\}$ We know $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$ Here, $a=1$ and $b=\cos ^{2} x$ Hence, $\int\left(1-\cos ^{2} x\right)^{3} \sin x d x=\int\left(1-\cos ^{6} x-3 \cos ^{2} x+3 \cos ^{4} x\right) \sin x d x$ $=\in...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int x \cos ^{3} x^{2} \sin x^{2} d x$ Solution: Let $\cos x^{2}=t$ Then $d\left(\cos x^{2}\right)=d t$ Since $d\left(x^{n}\right)=n x^{n-1}$ and $d(\cos x)=-\sin x d x$ $d t=2 x\left(-\sin x^{2}\right)=-2 x \sin x^{2} d x$ $x \sin x^{2} d x=-\frac{d t}{2}$ hence $\int x \cos ^{3} x^{2} \sin x^{2} d x=\int t^{3} x-\frac{d t}{2}$ $=-\frac{1}{2} \int \mathrm{t}^{3} \mathrm{dt}$ $=-\frac{1}{2} \times \frac{\mathrm{t}^{4}}{4}+\mathrm{c}$ $=-\frac{1}{8} \c...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \cos ^{7} x d x$ Solution: $\int \cos ^{7} x d x=\int \cos ^{6} x \cos x d x$ $=\int\left(\cos ^{2} x\right)^{3} \cos x d x$ $=\int\left(1-\sin ^{2} x\right)^{3} \cos x d x\left\{\right.$ since $\left.\sin ^{2} x+\cos ^{2} x=1\right\}$ We know $(a-b)^{3}=a^{3} b^{3}-3 a^{2} b+3 a b^{2}$ Here, $a=1$ and $b=\sin ^{2} x$ Hence, $\int\left(1-\sin ^{2} x\right)^{3} \cos x d x=\int\left(1-\sin ^{6} x-3 \sin ^{2} x+3 \sin ^{4} x\right) \cos x d x$ $=\in...
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