Find the value
Question: Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, When $\mathrm{y}=\cos \left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)$ Solution: Let $y=\cos \left(\frac{1-x^{2}}{1+x^{2}}\right), u=1-x^{2}, v=1+x^{2}, z=\frac{1-x^{2}}{1+x^{2}}$ Formula : $\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=2 \mathrm{x}$ and $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$ According to the quotient rule of differentiation If $z=\frac{u}{v}$ $\mathrm{dz} / \mathrm{dx}=\frac{\mathrm...
Read More →A test charge q is made to move in the electric
Question: A test charge q is made to move in the electric field of a point charge Q along two different closed paths. First path has sections along and perpendicular to lines of the electric field, the second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases? Solution: Work done is zero in both the cases. The work done by the electric force on the charge is in the closed-loop which is equal to zero....
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, $\int \frac{1}{\sqrt{a^{2}+b^{2} x^{2}}} d x=\frac{1}{b} \int \frac{1}{\sqrt{\left(a^{2}+t^{2}\right)}} d t$ $=\frac{1}{b} \log \left[t+\sqrt{a^{2}+t^{2}}\right]+c\left\{\right.$ since $\left.\left.\int \frac{1}{\sqrt{\left(a^{2}+x^{2}\right)}} d x=\log \left[x+\sqrt{\left(a^{2}\righ...
Read More →Can the potential function have a maximum
Question: Can the potential function have a maximum or minimum in free space? Solution: No, potential function cannot be maximum or minimum in free space as the absence of atmosphere around the conductor prevents the electric discharge....
Read More →Find the value
Question: Find $\frac{d y}{d x}$, When $=e^{x} \log (\sin 2 x)$ Solution: Let $y=e^{x} \log (\sin 2 x), z=e^{x}$ and $w=\log (\sin 2 x)$ Formula : $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}, \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}=\frac{1}{\mathrm{x}}$ and $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ According to the product rule of differentiation $\mathrm{dy} / \mathrm{dx}=\mathrm{w} \times \frac{\mathrm{dz}}{\mathrm...
Read More →Can there be a potential difference between
Question: Can there be a potential difference between two adjacent conductors carrying the same charge? Solution: Yes, there can be a potential difference between two adjacent conductors carrying the same charge....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{1+4 x^{2}}} d x$ Solution: Let $\mathrm{I}=\int \frac{1}{\sqrt{1+4 \mathrm{x}^{2}}} \mathrm{~d} \mathrm{x}=\int \frac{1}{\sqrt{1+(2 \mathrm{x})^{2}}} \mathrm{dx}$ Let $\mathrm{t}=2 \mathrm{x}$, then $\mathrm{dt}=2 \mathrm{dx}$ or $\mathrm{dx}=\mathrm{dt} / 2$ Therefore, $\int \frac{1}{\sqrt{1+(2 x)^{2}}} \mathrm{dx}=\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{1+t^{2}}}$ $=\frac{1}{2} \log \left[\mathrm{t}+\sqrt{1+\mathrm{t}^{2}}\righ...
Read More →Do free electrons travel to a region
Question: Do free electrons travel to a region of higher potential or lower potential? Solution: The force on the charged particle in the electric field is F = qE The direction of the electric field and the direction of electrostatic force experienced by the free electrons are in opposite directions The direction of the electric field is higher than the potential and therefore, the electrons travel from a lower potential region to higher potential....
Read More →Consider two conducting spheres of radii R1 and R2 with R1 > R2.
Question: Consider two conducting spheres of radii R1and R2with R1 R2. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. Solution: There are two spheres $\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}}{R_{1}}$ $\frac{1}{4 \pi \epsilon_{0}} \frac{q_{2}}{R_{2}}$ 1R1= 2R2 1/ 2= R2/R1 As R2 R1, that means 1 2 Therefore, the charge density of the smaller sphere ...
Read More →Find the value
Question: Find $\frac{d y}{d x}$, When $y=\sin \sqrt{\sin x+\cos x}$ Solution: Let $y=\sin (\sqrt{\sin x+\cos x}), z=\sqrt{\sin x+\cos x}$ Formula : $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$ and $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ $\frac{d(\sqrt{\sin x+\cos x})}{d x}=\frac{1}{2} \times(\sin x+\cos x)^{\frac{1}{2}-1} \times(\cos x-\sin x)$ According to the chain rule of differentiation $\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{2}-1}{x^{2}+4} d x$ Solution: Add and subtract 4 in the numerator, we get $=\int \frac{x^{2}+4-4-1}{x^{2}+4}=\int \frac{\left(x^{2}+4\right)-4-1}{x^{2}+4} d x$ $=\int \frac{\left(x^{2}+4\right)-5}{x^{2}+4} d x=\int \frac{\left(x^{2}+4\right)}{x^{2}+4} d x-\int \frac{5}{x^{2}+4} d x$ \{separate the numerator terms) $=\int d x-\int \frac{5}{x^{2}+4} d x=\int d x-5 \int \frac{1}{x^{2}+4} d x$ $=\int \mathrm{dx}-5 \int \frac{1}{\mathrm{x}^{2...
Read More →A parallel plate capacitor is connected to a battery
Question: A parallel plate capacitor is connected to a battery as shown in the figure. Consider two situations: A: Key K is kept closed and plates of capacitors are moved apart using the insulating handle B: Key K is opened and plates of capacitors are moved apart using the insulating handle Choose the correct options (a) In A: Q remains the same but C changes (b) In B: V remains the same but C changes (c) In A: V remains the same and hence Q changes (d) In B: Q remains the same and hence V chan...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t x: $\frac{e^{2 x}+x^{3}}{\operatorname{cosec} 2 x}$ Solution: 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 $\frac{\mathrm{d}(\operatorname{cosec} \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec} \mathrm{x} \cot \mathrm{x}$ According to the quotient rule of differentiation if $y=\frac{u}{v}$ $\math...
Read More →If a conductor has a potential V ≠ 0 and there are no charges
Question: If a conductor has a potential V 0 and there are no charges anywhere else outside, then (a) there must be charges on the surface or inside itself (b) there cannot be any charge in the body of the conductor (c) there must be charges only on the surface (d) there must be charges inside the surface Solution: The correct answer is (a) there must be charges on the surface or inside itself (b) there cannot be any charge in the body of the conductor...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{a^{2} x^{2}+b^{2}} d x$ Solution: take out $a^{2}$ $=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\frac{b^{2}}{a^{2}}} d x$ $=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\left(\frac{b}{a}\right)^{2}} d x=\frac{1}{a^{2}} * \frac{1}{\left(\frac{b}{a}\right)} \tan ^{-1}\left[\frac{x}{\frac{b}{a}}\right]+c\left\{\right.$ since $\left.\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{b}{a}\right)+c\right\}$ $=\frac{1}{\mathrm{ab}} \tan ^{-1}\le...
Read More →In the circuit shown in the figure,
Question: In the circuit shown in the figure, initially key K1is closed and K2is open. Then K1is opened and K2is closed. Then (a) charge on C1gets redistributed such that V1= V2 (b) charge on C1gets redistributed such that Q1 = Q2 (c) charge on C1gets redistributed such that C1V1+ C2V2= C1E (d) charge on C1gets redistributed such that Q1 + Q2 = Q Solution: The correct answer is (a) charge on C1gets redistributed such that V1= V2 (d) charge on C1gets redistributed such that Q1 + Q2 = Q...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t x: $\sqrt{\frac{1+e^{x}}{1-e^{x}}}$ Solution: Let $y=\sqrt{\frac{1+e^{x}}{1-e^{x}}}, u=1+e^{x}, v=1-e^{x}, z=\frac{1+e^{x}}{1-e^{x}}$ Formula : $\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$ According to the quotient rule of differentiation If $\mathrm{z}=\frac{\mathrm{u}}{\mathrm{v}}$ $\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\ma...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{a^{2} x^{2}-b^{2}} d x$ Solution: take out $a^{2}$ $=\frac{1}{a^{2}} \int \frac{1}{x^{2}-\frac{b^{2}}{a^{2}}} d x$ $=\frac{1}{a^{2}} \int \frac{1}{x^{2}-\left(\frac{b}{a}\right)^{2}} d x=\frac{1}{a^{2}} * \frac{1}{2\left(\frac{b}{a}\right)} \log \left[\frac{x-\left(\frac{b}{a}\right)}{x+\frac{b}{a}}\right]+c\left\{\operatorname{since} \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \frac{x+a}{x-a}+c\right\}$ $=\frac{1}{2 a b} \log \fra...
Read More →In a region of constant potential
Question: In a region of constant potential (a) the electric field is uniform (b) the electric field is zero (c) there can be no charge inside the region (d) the electric field shall necessarily change if a charge is placed outside the region Solution: The correct answer is (b) the electric field is zero (c) there can be no charge inside the region...
Read More →The work done to move a charge along
Question: The work done to move a charge along an equipotential from A to B (a) cannot be defined as $-\int_{A}^{B} E . d l$ (b) must be defined as $-\int_{A}^{B} E . d l$ (c) is zero (d) can have a non-zero value Solution: The correct answer is (b) must be defined as $-\int_{A}^{B} E . d l$ (c) is zero...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{a^{2}-b^{2} x^{2}} d x$ Solution: Taking out $b^{2}, \frac{1}{b^{2}} \int \frac{1}{\left(\frac{a^{2}}{b^{2}}\right)-x^{2}} d x$ $=\frac{1}{\mathrm{~b}^{2}} \int \frac{1}{\left(\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}\right)-\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{\mathrm{~b}^{2}} \int \frac{1}{\left(\frac{2}{\mathrm{~b}}\right)^{2}-\mathrm{x}^{2}} \mathrm{dx}$ $=\frac{1}{b^{2}} \times \frac{1}{2\left(\frac{2}{b}\right)} \log \left[\frac{\frac...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x$ Solution: $=x^{2}+2 x+10=x^{2}+2 x+1-1+10($ add and substract 1$)$ $\left.=\left(x^{2}+1\right)^{2}-1+10=x^{2}+1\right)^{2}+9$ $=\left(x^{2}+1\right)^{2}+3^{2}$ Put $x+1=t$ hence $d x=d t$ and $x=t-1$ $\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x=\int 1 /\left(\left(x^{2}+1\right)^{2}+3^{2}\right) d x$ $=\int \frac{1}{t^{2}+3^{2}} d t$ We have, $\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\mathrm{a}^{2}...
Read More →Equipotential surfaces
Question: Equipotential surfaces (a) are closer in regions of large electric fields compared to regions of lower electric fields (b) will be more crowded near sharp edges of a conductor (c) will be more crowded near regions of large charge densities (d) will always be equally spaced Solution: The correct answer is (a) are closer in regions of large electric fields compared to regions of lower electric fields (b) will be more crowded near sharp edges of a conductor (c) will be more crowded near r...
Read More →Differentiate w.r.t x:
Question: Differentiate w.r.t x: $\sqrt{\frac{1+\sin x}{1-\sin x}}$ Solution: Let $y=\sqrt{\frac{1+\sin x}{1-\sin x}}, u=1+\sin x, v=1-\sin x, z=\frac{1+\sin x}{1-\sin x}$ Formula : $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$ According to the quotient rule of differentiation If $\mathrm{z}=\frac{\mathrm{u}}{\mathrm{v}}$ $\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$ $=\fr...
Read More →Consider a uniform electric field in the
Question: Consider a uniform electric field in thedirection. The potential is a constant (a) in all space (b) for any x for a given z (c) for any y for a given z (d) on the x-y plane for a given z Solution: The correct answer is (b) for any x for a given z (c) for any y for a given z (d) on the x-y plane for a given z...
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