Very-Short-Answer Questions
Question: Very-Short-Answer Questions If the quadratic equation $p x^{2}-2 \sqrt{5} p x+15=0$ has two equal roots then find the value of $p$. Solution: It is given that the quadratic equation $p x^{2}-2 \sqrt{5} p x+15=0$ has two equal roots. $\therefore D=0$ $\Rightarrow(-2 \sqrt{5} p)^{2}-4 \times p \times 15=0$ $\Rightarrow 20 p^{2}-60 p=0$ $\Rightarrow 20 p(p-3)=0$ $\Rightarrow p=0$ or $p-3=0$ $\Rightarrow p=0$ or $p=3$ Forp= 0, we get 15 = 0, which is not true.p 0Hence, the value ofpis 3....
Read More →Very-Short-Answer Questions
Question: Very-Short-Answer Questions If the roots of the quadratic equation $2 x^{2}+8 x+k=0$ are equal then find the value of $k$. Solution: It is given that the roots of the quadratic equation $2 x^{2}+8 x+k=0$ are equal. $\therefore D=0$ $\Rightarrow 8^{2}-4 \times 2 \times k=0$ $\Rightarrow 64-8 k=0$ $\Rightarrow k=8$ Hence, the value ofkis 8....
Read More →The probability of a man hitting
Question: The probability of a man hitting a target is $\frac{1}{10}$. The least number of shots required, so that the probability of his hitting the target at least once is greater than $\frac{1}{4}$, is__________. Solution: $p=\frac{1}{10}, q=\frac{9}{10}$ $P($ not hitting target in $n$ trials $)=\left(\frac{9}{10}\right)^{n}$ $P($ at least one hit $)=1-\left(\frac{9}{10}\right)^{n}$ $\because 1-\left(\frac{9}{10}\right)^{n}\frac{1}{4} \Rightarrow(0.9)^{n}0.75$ $\therefore n_{\text {minimum }}...
Read More →The probability that a
Question: The probability that a randomly chosen 5 -digit number is made from exactly two digits is :(1) $\frac{135}{10^{4}}$(2) $\frac{121}{10^{4}}$(3) $\frac{150}{10^{4}}$(4) $\frac{134}{10^{4}}$Correct Option: 1 Solution: Total outcomes $=9\left(10^{4}\right)$ Favourable outcomes $={ }^{9} C_{2}\left(2^{5}-2\right)+{ }^{9} C_{1}\left(2^{4}-1\right)=36(30)+9(15)$ Probability $=\frac{36 \times 30+9 \times 15}{9 \times 10^{4}}=\frac{4 \times 30+15}{10^{4}}=\frac{135}{10^{4}}$...
Read More →A die is thrown two times and the sum
Question: A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4 . Then the conditional probability that the score 4 has appeared atleast once is :(1) $\frac{1}{4}$(2) $\frac{1}{3}$(3) $\frac{1}{8}$(4) $\frac{1}{9}$Correct Option: , 4 Solution: $E_{1}$ [the event for getting score a multiple of 4 ] $=(1,3),(3,1),(2,2),(2,6),(6,2),(3,5),(5,3),(4,4) \(6,6)$ $E_{2}[4$ has appeared atleast once $]$ $=(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),...
Read More →A beam of protons with speed
Question: A beam of protons with speed $4 \times 10^{5} \mathrm{~ms}^{-1}$ enters a uniform magnetic field of $0.3 \mathrm{~T}$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton $=1.67 \times 10^{-27} \mathrm{~kg}$, charge of the proton $=1.69 \times 10^{-19} \mathrm{C}$ )(1) $2 \mathrm{~cm}$(2) $5 \mathrm{~cm}$(3) $12 \mathrm{~cm}$(4) $4 \mathrm{~cm}$Correct Option: , 4 Solution: (4) Pitch $=(v \cos \theta) T$...
Read More →Magnetic fields at two points on the axis of a circular coil at a distance
Question: Magnetic fields at two points on the axis of a circular coil at a distance of $0.05 \mathrm{~m}$ and $02 \mathrm{~m}$ from the centre are in the rato $8: 1$. The radius of coil is(1) $0.15 \mathrm{~m}$(2) $0.2 \mathrm{~m}$(3) $0.1 \mathrm{~m}$(4) $1.0 \mathrm{~m}$Correct Option: , 3 Solution: $B=\frac{\mu_{0} N i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}$ at $\mathrm{x}_{1}=0.05 \mathrm{~m}, \mathrm{~B}_{1}=\frac{\mu_{0} \mathrm{NiR}^{2}}{2\left(\mathrm{R}^{2}+(0.05)^{2}\right)^{3 / 2}...
Read More →The pair that does NOT require calcination is :
Question: The pair that does NOT require calcination is :$\mathrm{ZnO}$ and $\mathrm{MgO}$$\mathrm{ZnO}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}$$\mathrm{ZnCO}_{3}$ and $\mathrm{CaO}$$\mathrm{Fe}_{2} \mathrm{O}_{3}$ and $\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3}$Correct Option: 1 Solution: Calcination involves the conversion of metal carbonates or hydrated oxides into metal oxides. $\mathrm{ZnO}$ and $\mathrm{MgO}$ are oxides, therefore, does not require calcination...
Read More →Solve the following
Question: Let $E^{C}$ denote the complement of an event $E$. Let $E_{1}, E_{2}$ and $E_{3}$ be any pairwise independent events with $P\left(E_{1}\right)0$ and $\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3}\right)=0$. Then $\mathrm{P}\left(\mathrm{E}_{2}^{\mathrm{C}} \cap \mathrm{E}_{3}^{\mathrm{C}} / \mathrm{E}_{1}\right)$ is equal to :(1) $P\left(E_{2}^{C}\right)+P\left(E_{3}\right)$(2) $\mathrm{P}\left(\mathrm{E}_{3}^{\mathrm{C}}\right)-\mathrm{P}\left(\mathrm{E}_{2}^{...
Read More →Very-Short-Answer Questions
Question: Very-Short-Answer Questions Find the solution of the quadratic equation $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$. Solution: The given quadratic equation is $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$. $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$ $\Rightarrow 3 \sqrt{3} x^{2}+9 x+x+\sqrt{3}=0$ $\Rightarrow 3 \sqrt{3} x(x+\sqrt{3})+1(x+\sqrt{3})=0$ $\Rightarrow(x+\sqrt{3})(3 \sqrt{3} x+1)=0$ $\Rightarrow x+\sqrt{3}=0$ or $3 \sqrt{3} x+1=0$ $\Rightarrow x=-\sqrt{3}$ or $x=-\frac{1}{3 \sqrt{3}}=-\frac{\sqrt{3}}{9}...
Read More →Given below are two statements :
Question: Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$. Assertion $A$ : When a rod lying freely is heated, no thermal stress is developed in it. Reason $\mathrm{R}$ : On heating, the length of the rod increases. In the light of the above statements, choose the corect answer from the options given below :(1) $\mathrm{A}$ is true but $\mathrm{R}$ is false(2) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$(3) Both $A$ an...
Read More →The reaction that does NOT define calcination is:
Question: The reaction that does NOT define calcination is: Correct Option: Solution: Calcination involves heating of carbonates or hydrated oxides in absence of air....
Read More →Very-Short-Answer Questions
Question: Very-Short-Answer Questions Find the roots of the quadratic equation $2 x^{2}-x-6=0$. Solution: The given quadratic equation is $2 x^{2}-x-6=0$. $2 x^{2}-x-6=0$ $\Rightarrow 2 x^{2}-4 x+3 x-6=0$ $\Rightarrow 2 x(x-2)+3(x-2)=0$ $\Rightarrow(x-2)(2 x+3)=0$ $\Rightarrow x-2=0$ or $2 x+3=0$ $\Rightarrow x=2$ or $x=-\frac{3}{2}$ Hence, the roots of the given equation are 2 and $-\frac{3}{2}$....
Read More →The time taken for the magnetic energy to reach 25 % of its maximum value,
Question: The time taken for the magnetic energy to reach $25 \%$ of its maximum value, when a solenoid of resistance $R$, inductance $L$ is connected to a battery, is :(1) $\frac{\mathrm{L}}{\mathrm{R}} \ell \mathrm{n} 5$(2) infinite(3) $\frac{\mathrm{L}}{\mathrm{R}} \ell \mathrm{n} 2$(4) $\frac{L}{R} \ell n 10$Correct Option: , 3 Solution: (3) Magnetic energy $=\frac{1}{2} \mathrm{Li}^{2}=25 \%$ $\mathrm{ME} \Rightarrow 25 \% \Rightarrow \mathrm{i}=\frac{\mathrm{i}_{0}}{2}$ $\mathrm{i}=\mathrm...
Read More →Very-Short-Answer Questions
Question: Very-Short-Answer Questions If $x=\frac{-1}{2}$ is a solution of the quadratic equation $3 x^{2}+2 k x-3=0$, find the value of $k$. Solution: It is given that $x=\frac{-1}{2}$ is a solution of the quadratic equation $3 x^{2}+2 k x-3=0$. $\therefore 3 \times\left(\frac{-1}{2}\right)^{2}+2 k \times\left(\frac{-1}{2}\right)-3=0$ $\Rightarrow \frac{3}{4}-k-3=0$ $\Rightarrow k=\frac{3-12}{4}=-\frac{9}{4}$ Hence, the value of $k$ is $-\frac{9}{4}$....
Read More →Box I contains 30 cards numbered
Question: Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50 . A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is :(1) $\frac{2}{3}$(2) $\frac{8}{17}$(3) $\frac{4}{17}$(4) $\frac{2}{5}$Correct Option: , 2 Solution: Let $B_{1}$ and $B_{2}$ be the boxes and $N$ be the number of non-prime number. and $P$ (non-prime number) $=P\left(B_{1}\right) \...
Read More →Match the ores (column A) with the metals (column B) :
Question: Match the ores (column A) with the metals (column B) : (I) - (a); (II) - (b); (III) - (c); (IV) - (d)(I) - (c); (II) - (d); (III) - (b); (IV) - (a)(I) - (c); (II) - (d); (III) - (a); (IV) - (b)(I) $-(b) ;(I I)-(c) ;(I I I)-(d) ;(I V)-(a)$Correct Option: , 2 Solution:...
Read More →Very-Short-Answer Questions
Question: Very-Short-Answer Questions Show that $x=-2$ is a solution of $3 x^{2}+13 x+14=0$. Solution: The given equation is $3 x^{2}+13 x+14=0$. Puttingx= 2 in the given equation, we get $\mathrm{LHS}=3 \times(-2)^{2}+13 \times(-2)+14=12-26+14=0=\mathrm{RHS}$ x= 2 is a solution of the given equation....
Read More →A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field.
Question: A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.(1) Loop assumes circular shape with its plane normal to the field.(2) Loop assumes circular shape with its plane parallel to the field.(3) Wire gets stretched to become straight.(4) Shape of the loop remains unchanged.Correct Option: 1 Solution: (1) Every part ( $\mathrm{d} \ell$ ) of the wire is pulled by force $\mathrm{i}(\mathrm{d} \ell) ...
Read More →A seven digit number is formed using
Question: A seven digit number is formed using digit $3,3,4,4,4,5,5$. The probability, that number so formed is divisible by 2 , is :(1) $\frac{6}{7}$(2) $\frac{4}{7}$(3) $\frac{3}{7}$(4) $\frac{1}{7}$Correct Option: , 3 Solution: $\mathrm{n}(\mathrm{s})=\frac{7 !}{21321}$ $\mathrm{n}(\mathrm{E})=\frac{6 !}{2 ! 2 ! 2 !}$ $\mathrm{P}(\mathbf{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6 !}{7 !} \times \frac{2 ! 312 !}{2 ! 2 ! 2 !}$ $\frac{1}{7} \times 3=\frac{3}{7}$...
Read More →Very-Short-Answer Questions
Question: Very-Short-Answer Questions Show that $x=-3$ is a solution of $x^{2}+6 x+9=0$. Solution: The given equation is $x^{2}+6 x+9=0$. Puttingx= 3 in the given equation, we get LHS $=(-3)^{2}+6 \times(-3)+9=9-18+9=0=$ RHS x= 3 is a solution of the given equation....
Read More →Hall-Heroult's process is given by:
Question: Hall-Heroult's process is given by:Correct Option: , 3 Solution: Hall Heroult's process is an electrochemical process used in extraction of Al from alumina....
Read More →Four identical long solenoids A, B, C and D
Question: Four identical long solenoids $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ are connected to each other as shown in the figure. If the magnetic field at the center of $\mathrm{A}$ is $3 \mathrm{~T}$ the field at the center of $\mathrm{C}$ would be : (Assume that the magnetic field is confined with in the volume of respective solenoid). (1) $12 \mathrm{~T}$(2) $6 \mathrm{~T}$(3) $9 \mathrm{~T}$(4) $1 \mathrm{~T}$Correct Option: , 4 Solution: (4) $\phi \propto \mathrm{i}$ $\Right...
Read More →A fair coin is tossed a fixed number of times.
Question: A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :(1) $\frac{15}{2^{12}}$(2) $\frac{15}{2^{13}}$(3) $\frac{15}{2^{14}}$(4) $\frac{15}{2^{8}}$Correct Option: , 2 Solution: $\mathrm{p}(\mathrm{x}=9)=\mathrm{p}(\mathrm{x}=7)$ ${ }^{\mathrm{n}} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{\mathrm{n}-9} \times\left(\frac{1}{2}\right)^{9}={ }^{\mathrm{n}} \mathrm{C}_{7}\left(...
Read More →The sum of two natural numbers is 8 and their product is 15
Question: The sum of two natural numbers is 8 and their product is 15. Find the numbers. Solution: Let the required natural numbers bexand (8 x).It is given that the product of the two numbers is 15. $\therefore x(8-x)=15$ $\Rightarrow 8 x-x^{2}=15$ $\Rightarrow x^{2}-8 x+15=0$ $\Rightarrow x^{2}-5 x-3 x+15=0$ $\Rightarrow x(x-5)-3(x-5)=0$ $\Rightarrow(x-5)(x-3)=0$ $\Rightarrow x-5=0$ or $x-3=0$ $\Rightarrow x=5$ or $x=3$ Hence, the required numbers are 3 and 5....
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