An electron is moving along +x direction with a velocity
Question: An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6} \mathrm{~ms}^{-1}$. It enters a region of uniform electric field of $300 \mathrm{~V} / \mathrm{cm}$ pointing along $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be :(1) $3 \times 10^{-4} \mathrm{~T}$, along $+z$ direction(2) $5 \times 10^{-3} \mathrm{~T}$, along $-z$ direction(3) $5 \times 10^{-3} \math...
Read More →Short-Answer Questions
Question: Short-Answer Questions Solve: $x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$ Solution: $x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$ $\Rightarrow x^{2}+6 x-(a+4)(a-2)=0$ $\Rightarrow x^{2}+[(a+4)-(a-2)] x-(a+4)(a-2)=0$ $\Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0$ $\Rightarrow x[x+(a+4)]-(a-2)[x+(a+4)]=0$ $\Rightarrow[x+(a+4)][x-(a-2)]=0$ $\Rightarrow x+(a+4)=0$ or $x-(a-2)=0$ $\Rightarrow x=-(a+4)$ or $x=(a-2)$ Hence, $-(a+4)$ and $(a-2)$ are the roots of the given equation....
Read More →The increasing order of the boiling point of the major products A,
Question: The increasing order of the boiling point of the major products $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ of the following reactions will be: $\mathrm{B}\mathrm{C}\mathrm{A}$$\mathrm{C}\mathrm{A}\mathrm{B}$$ABC$$\mathrm{A}\mathrm{C}\mathrm{B}$Correct Option: 1 Solution: The boiling points of isomeric haloalkanes decrease with increase in branching. So order of B.P. is A C B....
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Question: Short-Answer Questions Solve: $x^{2}+5 x-\left(a^{2}+a-6\right)=0$ Solution: $x^{2}+5 x-\left(a^{2}+a-6\right)=0$ $\Rightarrow x^{2}+5 x-(a+3)(a-2)=0$ $\Rightarrow x^{2}+[(a+3)-(a-2)] x-(a+3)(a-2)=0$ $\Rightarrow x^{2}+(a+3) x-(a-2) x-(a+3)(a-2)=0$ $\Rightarrow x[x+(a+3)]-(a-2)[x+(a+3)]=0$ $\Rightarrow[x+(a+3)][x-(a-2)]=0$ $\Rightarrow x+(a+3)=0$ or $x-(a-2)=0$ $\Rightarrow x=-(a+3)$ or $x=(a-2)$ Hence, $-(a+3)$ and $(a-2)$ are the roots of the given equation....
Read More →A particle of charge q and mass m
Question: A particle of charge $q$ and mass $m$ is moving with a velocity$v \hat{i}(v \neq 0)$ towards a large screen placed in the Y-Zplane at a distance $d$. If there is a magnetic field $\vec{B}=B_{0} \hat{k}$, the minimum value of $v$ for which the particle will not hit the screen is:(1) $\frac{q d B_{0}}{3 m}$(2) $\frac{2 q d B_{0}}{m}$(3) $\frac{q d B_{0}}{m}$(4) $\frac{q d B_{0}}{2 m}$Correct Option: , 3 Solution: (3) In uniform magnetic field particle moves in a circular path, if the rad...
Read More →Four persons can hit a target correctly
Question: Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is:(1) $\frac{25}{192}$(2) $\frac{7}{32}$(3) $\frac{1}{192}$(4) $\frac{25}{32}$Correct Option: , 4 Solution: $P$ (at least one hit the target) $=1-P$ (none of them hit the target) $=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(...
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Question: Short-Answer Questions Solve: $4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0$ Solution: $4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0$ $\Rightarrow 4 x^{2}+4 b x-(a-b)(a+b)=0$ $\Rightarrow 4 x^{2}+2[(a+b)-(a-b)] x-(a-b)(a+b)=0$ $\Rightarrow 4 x^{2}+2(a+b) x-2(a-b) x-(a-b)(a+b)=0$ $\Rightarrow 2 x[2 x+(a+b)]-(a-b)[2 x+(a+b)]=0$ $\Rightarrow[2 x+(a+b)][2 x-(a-b)]=0$ $\Rightarrow 2 x+(a+b)=0$ or $2 x-(a-b)=0$ $\Rightarrow x=-\frac{a+b}{2}$ or $x=\frac{a-b}{2}$ Hence, $-\frac{a+b}{2}$ and $\frac{a-...
Read More →The minimum number of times one has to toss
Question: The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least $90 \%$ is :(1) 5(2) 3(3) 4(4) 2Correct Option: , 3 Solution: Let, $p$ is probability for getting head and is probability for getting tail. $p=P(H)=\frac{1}{2}, q=1-p=\frac{1}{2}$ $P(x \geq 1) \geq \frac{9}{10} \Rightarrow 1-P(x=0) \geq \frac{9}{10}$ $1-{ }^{n} C_{0}\left(\frac{1}{2}\right)^{n} \geq \frac{9}{10} \Rightarrow \frac{1}{2^{n}} \leq 1-\frac{9}{10} \Righ...
Read More →The major product of the following reaction is :
Question: The major product of the following reaction is : Correct Option: , 4 Solution:...
Read More →Short-Answer Questions
Question: Short-Answer Questions Solve: $4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$ Solution: $4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$ $\Rightarrow 4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3}=0$ $\Rightarrow 4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0$ $\Rightarrow(\sqrt{3} x+2)(4 x-\sqrt{3})=0$ $\Rightarrow \sqrt{3} x+2=0$ or $4 x-\sqrt{3}=0$ $\Rightarrow x=-\frac{2}{\sqrt{3}}=-\frac{2 \sqrt{3}}{3}$ or $x=\frac{\sqrt{3}}{4}$ Hence, $-\frac{2 \sqrt{3}}{3}$ and $\frac{\sqrt{3}}{4}$ are the roots of the given equation....
Read More →A square loop of side 2 a, and carrying current I,
Question: A square loop of side $2 a$, and carrying current $I$, is kept in $\mathrm{XZ}$ plane with its centre at origin. A long wire carrying the same current $I$ is placed parallel to the $z$-axis and passing through the point $(0, b, 0),(ba)$. The magnitude of the torque on the loop about $z$-axis is given by:(1) $\frac{\mu_{0} I^{2} a^{2}}{2 \pi b}$(2) $\frac{\mu_{0} I^{2} a^{3}}{2 \pi b^{2}}$(3) $\frac{2 \mu_{0} I^{2} a^{2}}{\pi b}$(4) $\frac{2 \mu_{0} I^{2} a^{3}}{\pi b^{2}}$Correct Optio...
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Question: Short-Answer Questions Solve: $\sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$ Solution: $\sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$ $\Rightarrow \sqrt{3} x^{2}-3 \sqrt{2} x+\sqrt{2} x-2 \sqrt{3}=0$ $\Rightarrow \sqrt{3} x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})=0$ $\Rightarrow(x-\sqrt{6})(\sqrt{3} x+\sqrt{2})=0$ $\Rightarrow x-\sqrt{6}=0$ or $\sqrt{3} x+\sqrt{2}=0$ $\Rightarrow x=\sqrt{6}$ or $x=-\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{6}}{3}$ Hence, $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$ are the root...
Read More →Let A and B be two non-null events such that
Question: Let $A$ and $B$ be two non-null events such that $A \subset B$. Then, which of the following statements is always correct?(1) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$(2) $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A})$(3) $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \leq \mathrm{P}(\mathrm{A})$(4) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=1$Correct Option: , 2 Solution: $\because A \subset B ;$ so $A \cap B=A$ Now, $P\left(\fra...
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Question: Short-Answer Questions Solve: $\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$ Solution: $\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$ $\Rightarrow \sqrt{3} x^{2}+12 x-2 x-8 \sqrt{3}=0$ $\Rightarrow \sqrt{3} x(x+4 \sqrt{3})-2(x+4 \sqrt{3})=0$ $\Rightarrow(x+4 \sqrt{3})(\sqrt{3} x-2)=0$ $\Rightarrow x+4 \sqrt{3}=0$ or $\sqrt{3} x-2=0$ $\Rightarrow x=-4 \sqrt{3}$ or $x=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$ Hence, $-4 \sqrt{3}$ and $\frac{2 \sqrt{3}}{3}$ are the roots of the given equation....
Read More →A circular coil has moment of inertia
Question: A circular coil has moment of inertia $0.8 \mathrm{~kg} \mathrm{~m}^{2}$ around any diameter and is carrying current to produce a magnetic moment of $20 \mathrm{Am}^{2}$. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 \mathrm{~T}$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{\circ}$ will be :(1) $10...
Read More →Short-Answer Questions
Question: Short-Answer Questions Solve: $3 x^{2}+5 \sqrt{5} x-10=0$ Solution: $3 x^{2}+5 \sqrt{5} x-10=0$ $\Rightarrow 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-10=0$ $\Rightarrow 3 x(x+2 \sqrt{5})-\sqrt{5}(x+2 \sqrt{5})=0$ $\Rightarrow(x+2 \sqrt{5})(3 x-\sqrt{5})=0$ $\Rightarrow x+2 \sqrt{5}=0$ or $3 x-\sqrt{5}=0$ $\Rightarrow x=-2 \sqrt{5}$ or $x=\frac{\sqrt{5}}{3}$ Hence, $-2 \sqrt{5}$ and $\frac{\sqrt{5}}{3}$ are the roots of the given equation....
Read More →Which of the following compounds will form the precipitate with aq.
Question: Which of the following compounds will form the precipitate with aq. $\mathrm{AgNO}_{3}$ solution most readily? Correct Option: , 4 Solution: Ease of precipitation of $\mathrm{AgBr}$ depends upon the rate of formation of carbocation. Most stable carbocation due to $+R$ effect of $N$....
Read More →A random variable X has the following probability distribution:
Question: A random variable $X$ has the following probability distribution: $\begin{array}{lllllll}\mathrm{X} : 1 2 3 4 5\end{array}$ $\mathrm{P}(\mathrm{X}) \quad: \quad \mathrm{K}^{2} \quad 2 \mathrm{~K} \quad \mathrm{~K} \quad 2 \mathrm{~K} \quad 5 \mathrm{~K}^{2}$ Then, $\mathrm{P}(\mathrm{X}2)$ is equal to:(1) $\frac{7}{12}$(2) $\frac{1}{36}$(3) $\frac{1}{6}$(4) $\frac{23}{36}$Correct Option: , 4 Solution: $\sum P(K)=1 \Rightarrow 6 K^{2}+5 K=1$ $6 K^{2}+5 K-1=0$ $6 K^{2}+6 K-K-1=0$ $\Right...
Read More →Short-Answer Questions
Question: Short-Answer Questions Solve: $2 x^{2}+a x-a^{2}=0$ Solution: $2 x^{2}+a x-a^{2}=0$ $\Rightarrow 2 x^{2}+2 a x-a x-a^{2}=0$ $\Rightarrow 2 x(x+a)-a(x+a)=0$ $\Rightarrow(x+a)(2 x-a)=0$ $\Rightarrow x+a=0$ or $2 x-a=0$ $\Rightarrow x=-a$ or $x=\frac{a}{2}$ Hence, $-a$ and $\frac{a}{2}$ are the roots of the given equation....
Read More →Short-Answer Questions
Question: Short-Answer Questions Solve: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$ Solution: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$ $\Rightarrow x^{2}-\sqrt{3} x-x+\sqrt{3}=0$ $\Rightarrow x(x-\sqrt{3})-1(x-\sqrt{3})=0$ $\Rightarrow(x-\sqrt{3})(x-1)=0$ $\Rightarrow x-\sqrt{3}=0$ or $x-1=0$ $\Rightarrow x=\sqrt{3}$ or $x=1$ Hence, 1 and $\sqrt{3}$ are the roots of the given equation....
Read More →A wire A, bent in the shape of an arc of a circle,
Question: A wire $A$, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius $2 \mathrm{~cm}$ and another wire $B$, also bent in the shape of arc of a circle, carrying a current of $3 \mathrm{~A}$ and having radius of $4 \mathrm{~cm}$, are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and $B$ at the common centre $O$ is : (1) $4: 6$(2) $6: 4$(3) $2: 5$(4) $6: 5$Correct Option: , 4 Solution: (4) Given : $I_{A}=2 \mathrm{~A}, R_{...
Read More →If 10 different balls are to be placed
Question: If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :(1) $\frac{965}{2^{11}}$(2) $\frac{965}{2^{10}}$(3) $\frac{945}{2^{10}}$(4) None of theseCorrect Option: , 4 Solution: Total number of ways placing 10 different balls in 4 distinct boxes $=4^{10}$ Since, two of the 4 distinct boxes contains exactly 2 and 3 balls. Then, there are three cases to place exactly 2 and 3 balls in 2 of the 4 boxe...
Read More →The decreasing order of reactivity of the following
Question: The decreasing order of reactivity of the following organic molecules towards $\mathrm{AgNO}_{3}$ solution is : $(\mathrm{C})(\mathrm{D})(\mathrm{A})(\mathrm{B})$$(\mathrm{A})(\mathrm{B})(\mathrm{D})(\mathrm{C})$$(\mathrm{A})(\mathrm{B})(\mathrm{C})(\mathrm{D})$$(\mathrm{B})(\mathrm{A})(\mathrm{C})(\mathrm{D})$Correct Option: , 4 Solution: Given reaction is $S_{N}$ 1 reaction. In $S_{N}$ l reaction Rate of reaction $\propto$ Stability of $\mathrm{C}^{+}$...
Read More →A small bar magnet placed with its axis
Question: A small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06 \mathrm{~T}$ experiences a torque of $0.018 \mathrm{Nm}$. The minimum work required to rotate it from its stable to unstable equilibrium position is :(1) $6.4 \times 10^{-2} \mathrm{~J}$(2) $9.2 \times 10^{-3} \mathrm{~J}$(3) $7.2 \times 10^{-2} \mathrm{~J}$(4) $11.7 \times 10^{-3} \mathrm{~J}$Correct Option: , 3 Solution: (3) Here, $\theta=30^{\circ}, \tau=0.018 \mathrm{~N}-\mathrm{m}, B=0.06 \math...
Read More →In a box, there are 20 cards,
Question: In a box, there are 20 cards, out of which 10 are labelled as $\mathrm{A}$ and the remaining 10 are labelled as $\mathrm{B}$. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third B-card is :(1) $\frac{9}{16}$(2) $\frac{11}{16}$(3) $\frac{13}{16}$(4) $\frac{15}{16}$Correct Option: , 2 Solution: $P($ second $A$ - card appears before the third $B$ - card) $=P(A A)+P(A B A)+P(B...
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