The value of
Question: The value of $\lim _{h \rightarrow 0}\left\{\frac{\left.\sqrt{3} \sin \left(\frac{\pi}{6}+h\right)^{-\cos \left(\frac{\pi}{6}+h\right.}\right)}{\sqrt{3} h(\sqrt{3} \cosh -\sinh )}\right\}$ is:(1) $\frac{3}{4}$(2) $\frac{2}{\sqrt{3}}$(3) $\frac{4}{3}$(4) $\frac{2}{3}$Correct Option: , 3 Solution: $\lim _{h \rightarrow 0} 2 \times 2\left\{\frac{\sin \left(\frac{\pi}{6}+h-\frac{\pi}{6}\right)}{2 \sqrt{3 h}\left(\cos \left(h+\frac{\pi}{6}\right)\right.}\right\}$ $=\frac{2}{\sqrt{3}} \times...
Read More →if
Question: If $\lim _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{\operatorname{ax}\left(e^{4 z}-1\right)}$ exists and is equal to $b$, then the value of $a-2 b$ is Solution: $\lim _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{\operatorname{ax}\left(e^{4 x}-1\right)}$ Applying L' Hospital Rule $\lim _{x \rightarrow 0} \frac{a-4 e^{4 x}}{a\left(e^{4 x}-1\right)+\operatorname{ax}\left(4 e^{4 x}\right)} \quad$ So $a=4$ Applying L' Hospital Rule $\lim _{x \rightarrow 0} \frac{-16 e^{4 z...
Read More →A particle of mass m and charge q is released from rest in a uniform electric field.
Question: A particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)Correct Option: , 2 Solution: Using $v^{2}-u^{2}=2 a S$ ...(1) Here, $u=0, s=x$ Also, $F_{\text {electric }}=m a$ Substituting the values in (i) we get $v^{2}=\frac{2 q E}{m} \cdot x$ $\Rightarrow q E=m a$ $\Rightarrow a=\...
Read More →Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
Question: Find the roots of each of the following equations, if they exist, by applying the quadratic formula: $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ Solution: The given equation is $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$. Comparing it with $a x^{2}+b x+c=0$, we get $a=\sqrt{2}, b=7$ and $c=5 \sqrt{2}$ $\therefore$ Discriminant, $D=b^{2}-4 a c=(7)^{2}-4 \times \sqrt{2} \times 5 \sqrt{2}=49-40=90$ So, the given equation has real roots. Now, $\sqrt{D}=\sqrt{9}=3$ $\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\f...
Read More →is equal to :
Question: $\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots \ldots++\frac{1}{n}}{n^{2}}\right)^{n}$ is equal to :(1) $\frac{1}{2}$(2) $\frac{1}{\mathrm{e}}$(3) 1(4) 0Correct Option: , 3 Solution: It is $1^{\infty}$ form $\mathrm{L}=\mathrm{e}^{\lim _{z \rightarrow \infty}}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots+\frac{1}{n}}{n}\right)$ $S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\ldots \ldots \l...
Read More →if
Question: If $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{3 x^{3}}$ is equal to $L$, then the value of $(6 \mathrm{~L}+1)$ is(1) $\frac{1}{6}$(2) $\frac{1}{2}$(3) 6(4) 2Correct Option: , 4 Solution: $\lim _{x \rightarrow 0} \frac{\left(x+\frac{x^{3}}{3 !} \cdots\right)-\left(x-\frac{x^{3}}{3} \cdots\right)}{3 x^{3}}=\frac{1}{6}$ So $6 L+1=2$...
Read More →Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
Question: Find the roots of each of the following equations, if they exist, by applying the quadratic formula: $2 x^{2}-2 \sqrt{2} x+1=0$ Solution: The given equation is $2 x^{2}-2 \sqrt{2} x+1=0$. Comparing it with $a x^{2}+b x+c=0$, we get $a=2, b=-2 \sqrt{2}$ and $c=1$ $\therefore$ Discriminant, $D=b^{2}-4 a c=(-2 \sqrt{2})^{2}-4 \times 2 \times 1=8-8=0$ So, the given equation has real roots. Now, $\sqrt{D}=0$ $\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{2})+\sqrt{0}}{2 \times ...
Read More →The value of the
Question: The value of the $\operatorname{limit} \lim _{\theta \rightarrow 0} \frac{\tan \left(\pi \cos ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$ is equal to :(1) $-\frac{1}{2}$(2) $-\frac{1}{4}$(3) 0(4) $\frac{1}{4}$Correct Option: 1 Solution: $\lim _{\theta \rightarrow 0} \frac{\tan \left(\pi\left(1-\sin ^{2} \theta\right)\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$ $=\lim _{\theta \rightarrow 0} \frac{-\tan \left(\pi \sin ^{2} \theta\right)}{\sin \left(2 \pi \sin ^...
Read More →Prove that
Question: $15 x^{2}-28=x$ Solution: Given: $15 x^{2}-28=x$ $\Rightarrow 15 x^{2}-x-28=0$ On comparing it with $a x^{2}+b x+c=0$, we get: $a=15, b=-1$ and $c=-28$ Discriminant $D$ is given by: $D=\left(b^{2}-4 a c\right)$ $=(-1)^{2}-4 \times 15 \times(-28)$ $=1-(-1680)$ $=1+1680$ $=1681$ $=16810$ Hence, the roots of the equation are real. Roots $\alpha$ and $\beta$ are given by : $\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-1)+\sqrt{1681}}{2 \times 15}=\frac{1+41}{30}=\frac{42}{30}=\frac{7}{5}$ $\bet...
Read More →Solve this
Question: $16 x^{2}=24 x+1$ Solution: Given : $16 x^{2}=24 x+1$ $\Rightarrow 16 x^{2}-24 x-1=0$ On comparing it with $a x^{2}+b x+c=0$ $a=16, b=-24$ and $c=-1$ Discriminant $D$ is given by: $D=\left(b^{2}-4 a c\right)$ $=(-24)^{2}-4 \times 16 \times(-1)$ $=576+(64)$ $=6400$ Hence, the roots of the equation are real, Roots $\alpha$ and $\beta$ are given by : $\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-24)+\sqrt{640}}{2 \times 16}=\frac{24+8 \sqrt{10}}{32}=\frac{8(3+\sqrt{10})}{32}=\frac{(3+\sqrt{10}...
Read More →The value of
Question: The value of $\lim _{n \rightarrow \infty} \frac{[\mathrm{r}]+[2 \mathrm{r}]+\ldots \ldots+[\mathrm{nr}]}{\mathrm{n}^{2}}$, where $\mathrm{r}$ is non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$, is equal to :(1) $\frac{\mathrm{r}}{2}$(2) $\mathrm{r}$(3) $2 \mathrm{r}$(4) 0Correct Option: 1 Solution: We know that $\mathrm{r} \leq[\mathrm{r}]\mathrm{r}+1$ and $2 r \leq[2 r]2 r+1$ $3 r \leq[3 r]3 r+1$ Now, $\lim _{n \rightarrow \infty} \frac{n(n+1) \c...
Read More →Consider two charged metallic spheres
Question: Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of radii $R_{1}$ and $R_{2}$, respectively. The electric fields $E_{1}$ (on $S_{1}$ ) and $E_{2}$ (on $S_{2}$ ) on their surfaces are such that $E_{1} / E_{2}=R_{1} / R_{2}$. Then the ratio $V_{1}\left(\right.$ on $\left.S_{1}\right) / V_{2}\left(\right.$ on $\left.S_{2}\right)$ of the electrostatic potentials on each sphere is:(1) $R_{1} / R_{2}$(2) $\left(R_{1} / R_{2}\right)^{2}$(3) $\left(R_{2} / R_{1}\right)$(4) $\left(\fra...
Read More →The value of
Question: The value of $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}}$, where $[x]$ denotes the greatest integer $\leq x$ is :(1) $\pi$(2) 0(3) $\frac{\pi}{4}$(4) $\frac{\pi}{2}$Correct Option: , 4 Solution: $\left.\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1} x}{\left(1-x^{2}\right)} \times \frac{\sin ^{-1} x}{x}=\frac{\pi}{2}\right\}$...
Read More →Solve this
Question: $25 x^{2}+30 x+7=0$ Solution: Given: $25 x^{2}+30 x+7=0$ On comparing it with $a x^{2}+b x+c=0$, we get: $a=25, b=30$ and $c=7$ Discriminant $D$ is given by : $D=\left(b^{2}-4 a c\right)$ $=30^{2}-4 \times 25 \times 7$ $=900-700$ $=200$ $=2000$ Hence, the roots of the equation are real. Roots $\alpha$ and $\beta$ are given by : $\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-30+\sqrt{200}}{2 \times 25}=\frac{-30+10 \sqrt{2}}{50}=\frac{10(-3+\sqrt{2})}{50}=\frac{(-3+\sqrt{2})}{5}$ $\beta=\frac{-...
Read More →if
Question: If $\lim _{x \rightarrow 0} \frac{a e^{x}-b \cos x+c e^{-x}}{x \sin x}=2$, then $a+b+c$ is equal to_________. Solution: $\lim _{x \rightarrow 0} \frac{a e^{x}-b_{\cos x+c e^{-x}}}{x \sin x}=2$ $\Rightarrow \lim _{x \rightarrow 0} \frac{a\left(1+x+\frac{x^{2}}{2 !} \ldots\right)^{-b}\left(1-\frac{x^{2}}{2 !}+\ldots\right)+c\left(1-x+\frac{x^{2}}{2 !}\right)}{\left(\frac{x \sin x}{x}\right)^{x}}=2$ $a-b+c=0$ $a-c=0$ $\ \frac{a+b+c}{2}=2$ $\Rightarrow a+b+c=4$...
Read More →In finding the electric field using Gauss law the formula
Question: In finding the electric field using Gauss law the formula $|\vec{E}|=\frac{q_{\text {enc }}}{\epsilon_{0}|A|}$ is applicable. In the formula $\epsilon_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $q_{\text {enc }}$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?(1) Only when the Gaussian surface is an equipotential surface. Only when the Gaussian surface is an(2) equipotential surface and $|\vec{E...
Read More →Three charged particles A, B and C with charges
Question: Three charged particles $A, B$ and $C$ with charges $-4 q, 2 q$ and $-2 q$ are present on the circumference of a circle of radius $d$. The charged particles $A, C$ and centre $O$ of the circle formed an equilateral triangle as shown in figure. Electric field at $O$ along $x$-direction is: (1) $\frac{\sqrt{3 q}}{\pi \in_{0} d^{2}}$(2) $\frac{2 \sqrt{3 q}}{\pi \in_{0} d^{2}}$(3) $\frac{\sqrt{3 q}}{4 \pi \in_{0} d^{2}}$(4) $\frac{3 \sqrt{3 q}}{4 \pi \in_{0} d^{2}}$Correct Option: 1 Soluti...
Read More →All x satisfying the inequality
Question: All $x$ satisfying the inequality $\left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+100$, lie in the interval:(1) $(-\infty, \cot 5) \cup(\cot 4, \cot 2)$(2) $(\cot 2, \infty)$(3) $(-\infty, \cot 5) \cup(\cot 2, \infty)$(4) $(\cot 5, \cot 4)$Correct Option: , 2 Solution: $\left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+100$ $\left(\cot ^{-1} x-5\right)\left(\cot ^{-1}-2\right)0$ $\cot ^{-1} x \in(-\infty, 2) \cup(5, \infty)$ ...(1) But $\cot ^{-1} x$ lies in $(0, \pi)$...
Read More →The value of
Question: The value of $\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right)$ is: (1) $\frac{21}{19}$(2) $\frac{19}{21}$(3) $\frac{22}{23}$(4) $\frac{23}{22}$Correct Option: 1 Solution: $\operatorname{cat}\left(\sum_{n=1}^{19} \cot ^{-1}\left(1+\sum_{p=1}^{n} 2 p\right)\right)$ $=\cot \left(\sum_{n=1}^{19} \cot ^{-1}(1+n(n+1))\right)$ $=\cot \left(\sum_{n=1}^{19} \tan ^{-1}\left(\frac{(n+1)-n}{1+(n+1) n}\right)\right)$ $\left[\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\righ...
Read More →Two infinite planes each with uniform surface charge density
Question: Two infinite planes each with uniform surface charge density $+\sigma$ are kept in such a way that the angle between them is $30^{\circ}$. The electric field in the region shown between them is given by: (1) $\frac{\sigma}{2 \in_{0}}\left[(1+\sqrt{3}) \hat{y}-\frac{\hat{x}}{2}\right]$(2) $\frac{\sigma}{\epsilon_{0}}\left[\left(1+\frac{\sqrt{3}}{2}\right) \hat{y}+\frac{\hat{x}}{2}\right]$(3) $\frac{\sigma}{2 \in_{0}}\left[(1+\sqrt{3}) \hat{y}+\frac{\hat{x}}{2}\right]$(4) $\frac{\sigma}{...
Read More →then y-x is equal to.
Question: If $x=\sin ^{-1}(\sin 10)$ and $y=\cos ^{-1}(\cos 10)$, then $y-x$ is equal to.(1) 0(2) 10(3) $7 \pi$(4) $\pi$Correct Option: , 4 Solution: $x=\sin ^{-1}(\sin 10)$ $\Rightarrow \quad x=3 \pi-10$ $\left\{\begin{array}{l}3 \pi-\frac{\pi}{2}103 \pi+\frac{\pi}{2} \\ \Rightarrow 3 \pi-x=10\end{array}\right.$ and $y=\cos ^{-1}(\cos 10)$ $\left\{\begin{array}{l}3 \pi104 \pi \\ \Rightarrow 4 \pi-x=10\end{array}\right.$ $\Rightarrow \quad y=4 \pi-10$ $\therefore \quad y-x=(4 \pi-10)-(3 \pi-10)=...
Read More →then x is equal to:
Question: If $\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2}\left(x\frac{3}{4}\right)$, then $x$ is equal to:(1) $\frac{\sqrt{145}}{12}$(2) $\frac{\sqrt{145}}{10}$(3) $\frac{\sqrt{146}}{12}$(4) $\frac{\sqrt{145}}{11}$Correct Option: 1 Solution: $\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2} ;\left(x\frac{3}{4}\right)$ $\Rightarrow \quad \cos ^{-1}\left(\frac{2}{3 x}\right)=\frac{\pi}{2}-\cos ^{-1}\left(\frac{3}{4 ...
Read More →Consider the force F on a charge ' q'
Question: Consider the force $\mathrm{F}$ on a charge ' $q$ ' due to a uniformly charged spherical shell of radius $R$ carrying charge $Q$ distributed uniformly over it. Which one of the following statements is true for F, if ' $q$ ' is placed at distance $r$ from the centre of the shell?(1) $\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{R^{2}}$ for $rR$(2) $\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{R^{2}}\mathrm{F}0$ for $rR$(3) $\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{R^...
Read More →The value of
Question: The value of $\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{3}{5}\right)$ is equal to :(1) $\pi-\sin ^{-1}\left(\frac{63}{65}\right)$(2) $\frac{\pi}{2}-\sin ^{-1}\left(\frac{56}{65}\right)$(3) $\frac{\pi}{2}-\cos ^{-1}\left(\frac{9}{65}\right)$(4) $\pi-\cos ^{-1}\left(\frac{33}{65}\right)$Correct Option: , 2 Solution: $-\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{12}{13}\right)=-\sin ^{-1}\left(\frac{3}{5} \times \frac{5}{13}-\frac{12}{13} \times \frac{4}{5}\r...
Read More →if
Question: If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, where $-1 \leq x \leq 1,-2 \leq y \leq 2$, $x \leq \frac{y}{2}$, then for all $x, y, 4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to:If $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$, where $-1 \leq x \leq 1,-2 \leq y \leq 2$, $x \leq \frac{y}{2}$, then for all $x, y, 4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to:(1) $4 \sin ^{2} \alpha$(2) $2 \sin ^{2} \alpha$(3) $4 \sin ^{2} \alpha-2 x^{2} y^{2}$(d) $4 \cos ^{2} \alpha+2 x^{2} y^{2}$Correct ...
Read More →