In the figure shown, what is the current (in Ampere) drawn from the battery? You are given :
Question: In the figure shown, what is the current (in Ampere) drawn from the battery? You are given : $\mathrm{R}_{1}=15 \Omega, \mathrm{R}_{2}=10 \Omega, \mathrm{R}_{3}=20 \Omega, \mathrm{R}_{4}=5 \Omega, \mathrm{R}_{5}=25 \Omega, \mathrm{R}_{6}$ $=30 \Omega, \mathrm{E}=15 \mathrm{~V}$ (1) $13 / 24$(2) $7 / 18$(3) $9 / 32$(4) $20 / 3$Correct Option: , 3 Solution: (3) $R_{3}, R_{4}$ and $R_{5}$ are in series so their equivalent $R=20+5+25=50 \Omega$ This is parallel with $R_{2}$, and so net res...
Read More →Find the zeros of the quadratic polynomial
Question: Find the zeros of the quadratic polynomial $\left(8 x^{2}-4\right)$ and verify the relation between the zeros and the coefficients. Solution: We have: $f(x)=8 x^{2}-4$ It can be written as $8 x^{2}+0 x-4$ $=4\left\{(\sqrt{2} x)^{2}-(1)^{2}\right\}$ $=4(\sqrt{2} x+1)(\sqrt{2} x-1)$ $\therefore f(x)=0=(\sqrt{2} x+1)(\sqrt{2} x-1)=0$ $=\sqrt{2} x+1=0$ or $\sqrt{2} x-1=0$ $\Rightarrow x=\frac{-1}{\sqrt{2}}$ or $x=\frac{1}{\sqrt{2}}$ So, the zeroes of $f(x)$ are $\frac{-1}{\sqrt{2}}$ and $\...
Read More →Let
Question: Let $\mathrm{f}: \mathrm{S} \rightarrow \mathrm{S}$ where $\mathrm{S}=(0, \infty)$ be a twice differentiable function such that $\mathrm{f}(\mathrm{x}+1)=\mathrm{xf}(\mathrm{x})$ If $g: S \rightarrow R$ be defined as $g(x)=\log _{e} f(x)$, then the value of $\left|g^{\prime \prime}(5)-\mathrm{g}^{\prime \prime}(1)\right|$ is equal to :(1) $\frac{205}{144}$(2) $\frac{197}{144}$(3) $\frac{187}{144}$(4) 1Correct Option: 1 Solution: $\operatorname{lnf}(x+1)=\ln (x f(x))$ $\operatorname{lnf...
Read More →Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question: Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients: $5 x^{2}+10 x$ Solution: Let $f(x)=5 x^{2}+10 x$ To find the zeroes, we put $f(x)=0$ $\Rightarrow 5 x^{2}+10 x=0$ $\Rightarrow 5 x(x+2)=0$ $\Rightarrow(x)(x+2)=0$ $\Rightarrow(x)=0$ or $(x+2)=0$ $\Rightarrow x=0,-2$ Hence, all the zeroes of the polynomial $f(x)$ are 0 and $-2$. Now, $f(0)=5(0)^{2}+10(0)$ $=0+0$ $=0$ $f(-2)=5(-2)^{2}+10(-2)$ $=5(4)-20$ $=20-20$ $=0$ ...
Read More →A cell of internal resistance r drives current through an external resistance R.
Question: A cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when :(1) $\mathrm{R}=0.001 r$(2) $\mathrm{R}=1000 \mathrm{r}$(3) $\mathrm{R}=2 r$(4) $\mathrm{R}=r$Correct Option: , 4 Solution: (4) $i=\left(\frac{\varepsilon}{R+r}\right)$ Power delivered to $\mathrm{R}$. $P=i^{2} R=\left(\frac{\varepsilon}{R+r}\right)^{2} R$ $\mathrm{P}$ to be maximum, $\frac{d P}{d R}=0$ or $\frac{d}{d R}\l...
Read More →Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question: Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients: $3 x^{2}-x-4$ Solution: Let $f(x)=3 x^{2}-x-4$ To find the zeroes, we put $f(x)=0$ $\Rightarrow 3 x^{2}-x-4=0$ $\Rightarrow 3 x^{2}-4 x+3 x-4=0$ $\Rightarrow x(3 x-4)+1(3 x-4)=0$ $\Rightarrow(x+1)(3 x-4)=0$ $\Rightarrow(x+1)=0$ or $(3 x-4)=0$ $\Rightarrow x=-1, \frac{4}{3}$ Hence, all the zeroes of the polynomial $f(x)$ are $-1$ and $\frac{4}{3}$. Now, $f(-1)=3(-1)^...
Read More →In the circuit shown, a four-wire potentiometer is made
Question: In the circuit shown, a four-wire potentiometer is made of a $400 \mathrm{~cm}$ long wire, which extends between $\mathrm{A}$ and $\mathrm{B}$. The resistance per unit length of the potentiometer wire is $r=0.01 \Omega / \mathrm{cm}$. If an ideal voltmeter is connected as shown with jockey $\mathrm{J}$ at $50 \mathrm{~cm}$ from end $\mathrm{A}$, the expected reading of the voltmeter will be : (1) $0.50 \mathrm{~V}$(2) $0.75 \mathrm{~V}$(3) $0.25 \mathrm{~V}$(4) $0.20 \mathrm{~V}$Correc...
Read More →Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
Question: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients $4 x^{2}-4 x+1$ Solution: $4 x^{2}-4 x+1=0$ $\Rightarrow(2 x)^{2}-2(2 x)(1)+(1)^{2}=0$ $\Rightarrow(2 x-1)^{2}=0 \quad\left[\because a^{2}-2 a b+b^{2}=(a-b)^{2}\right]$ $\Rightarrow(2 x-1)^{2}=0$ $\Rightarrow x=\frac{1}{2}$ or $x=\frac{1}{2}$ Sum of zeroes $=\frac{1}{2}+\frac{1}{2}=1=\frac{1}{1}=\frac{-(\text { coefficient of } x)}{\text { coefficient of } x^{2}}$ ...
Read More →For the circuit shown,
Question: For the circuit shown, with $R_{1}=1.0 \Omega, R_{2}=2.0 \Omega, E_{1}=2 \mathrm{~V}$ and $E_{2}=E_{3}=4 \mathrm{~V}$, the potential difference between the points ' $a$ ' and ' $b$ ' is approximately (in V) : (1) $2.7$(2) $2.3$(3) $3.7$(4) $3.3$Correct Option: , 4 Solution: (4) Applying parallel combination of batteries Equivalent emf between $\mathrm{A}$ and $\mathrm{B}$ is $=\frac{\frac{E_{1}}{1+1}+\frac{E_{2}}{2}+\frac{E_{3}}{1+1}}{\frac{1}{1+1}+\frac{1}{2}+\frac{1}{1+1}}$ $\frac{\f...
Read More →Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
Question: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients $2 \sqrt{3} x^{2}-5 x+\sqrt{3}$ Solution: $2 \sqrt{3} x^{2}-5 x+\sqrt{3}$ $\Rightarrow 2 \sqrt{3} x^{2}-2 x-3 x+\sqrt{3}$ $\Rightarrow 2 x(\sqrt{3} x-1)-\sqrt{3}(\sqrt{3} x-1)=0$ $\Rightarrow(\sqrt{3} x-1)$ or $(2 x-\sqrt{3})=0$ $\Rightarrow(\sqrt{3} x-1)=0$ or $(2 x-\sqrt{3})=0$ $\Rightarrow x=\frac{1}{\sqrt{3}}$ or $x=\frac{\sqrt{3}}{2}$ $\Rightarrow x=\frac{1}{\...
Read More →Find the zeros of the quadratic polynomial
Question: Find the zeros of the quadratic polynomial $2 x^{2}-11 x+15$ and verify the relation between the zeros and the coefficients. Solution: We have: $f(x)=2 x^{2}-11 x+15$ $=2 x^{2}-(6 x+5 x)+15$ $=2 x^{2}-6 x-5 x+15$ $=2 x(x-3)-5(x-3)$ $=(2 x-5)(x-3)$ $\therefore f(x)=0=(2 x-5)(x-3)=0$ $=2 x-5=0$ or $x-3=0$ $=x=\frac{5}{2}$ or $x=3$ So, the zeroes of $f(x)$ are $\frac{5}{2}$ and 3 . Sum of the zeroes $=\frac{5}{2}+3=\frac{5+6}{2}=\frac{11}{2}=\frac{-(\text { coefficient of } x)}{\left(\tex...
Read More →Find the zeros of the quadratic polynomial
Question: Find the zeros of the quadratic polynomial $5 x^{2}-4-8 x$ and verify the relationship between the zeros and the coefficients of the given polynomial. Solution: We have: $f(x)=5 x^{2}-4-8 x$ $=5 x^{2}-8 x-4$ $=5 x^{2}-(10 x-2 x)-4$ $=5 x^{2}-10 x+2 x-4$ $=5 x(x-2)+2(x-2)$ $=(5 x+2)(x-2)$ $\therefore f(x)=0=(5 x+2)(x-2)=0$ $=5 x+2=0$ or $x-2=0$ $=x=\frac{-2}{5}$ or $x=2$ So, the zeros of $f(x)$ are $\frac{-2}{5}$ and 2 . Sum of the zeros $=\left(\frac{-2}{5}\right)+2=\frac{-2+10}{5}=\fr...
Read More →Solve this
Question: A $200 \Omega$ resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be :(1) $100 \Omega$(2) $400 \Omega$(3) $300 \Omega$(4) $500 \Omega$Correct Option: , 4 Solution: (4) Number 2 is associated with the red colour. This colour is replaced by green. $\because$ Colour code figure for green is 5 $\therefore$ New resistance $=500 \Omega$...
Read More →Let y=y(x) be the solution of the differential equation,
Question: Let $y=y(x)$ be the solution of the differential equation, $x \frac{d y}{d x}+y=x \log _{\mathrm{e}} \mathrm{x},(\mathrm{x}1) .$ If $2 \mathrm{y}(2)=\log _{\mathrm{e}} 4-1$, then $\mathrm{y}(\mathrm{e})$ is equal to:(1) $-\frac{e}{2}$(2) $-\frac{e^{2}}{2}$(3) $\frac{e}{4}$(4) $\frac{e^{2}}{4}$Correct Option: , 3 Solution: Consider the differential equation, $\frac{d y}{d x}+\frac{y}{x}=\log _{e} x$ $\because \quad I F=e^{\int \frac{1}{x} d x}=x$ $\therefore \quad y x=\int x \ln x d x$ ...
Read More →Find the zeros of the quadratic polynomial
Question: Find the zeros of the quadratic polynomial $4 x^{2}-4 x-3$ and verify the relation between the zeros and the coefficients. Solution: We have: $f(x)=4 x^{2}-4 x-3$ $=4 x^{2}-(6 x-2 x)-3$ $=4 x^{2}-6 x+2 x-3$ $=2 x(2 x-3)+1(2 x-3)$ $=(2 x+1)(2 x-3)$ $\therefore f(x)=0=(2 x+1)(2 x-3)=0$ $=2 x+1=0$ or $2 x-3=0$ $=x=\frac{-1}{2}$ or $x=\frac{3}{2}$ So, the zeros of $f(x)$ are $\frac{-1}{2}$ and $\frac{3}{2}$. Sum of the zeros $=\left(\frac{-1}{2}\right)+\frac{3}{2}=\frac{-1+3}{2}=\frac{2}{2...
Read More →In a meter bridge experiment S is a standard resistance.
Question: In a meter bridge experiment $\mathrm{S}$ is a standard resistance. $R$ is a resistance wire. It is found that balancing length is $l=25 \mathrm{~cm}$. If $\mathrm{R}$ is replaced by a wire of half length and half diameter that of $\mathrm{R}$ of same material, then the balancing distance $l^{\prime}$ (in $\mathrm{cm}$ ) will now be Solution: (40) For the given meter bridge $\frac{R}{S}=\frac{\ell_{1}}{100-\ell_{1}}$ Where, $\ell_{1}=$ balancing length $\Rightarrow \quad \frac{R}{S}=\f...
Read More →The solution of the differential equation,
Question: The solution of the differential equation, $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^{2}$, when $y(1)=1$, is :(1) $\log _{\mathrm{e}}\left|\frac{2-x}{2-y}\right|=x-y$(2) $-\log _{\mathrm{e}}\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$(3) $-\log _{\mathrm{e}}\left|\frac{1+x-y}{1-x+y}\right|=x+y-2$(4) $\log _{\mathrm{e}}\left|\frac{2-y}{2-x}\right|=2(y-1)$Correct Option: , 2 Solution: The given differential equation $\frac{d y}{d x}=(x-y)^{2}$ .......(1) Let $x-y=t \Rightarrow 1-\frac{d y}{d...
Read More →Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:
Question: Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients: $2 x^{2}-x-6$ Solution: Let $f(x)=2 x^{2}-x-6$ To find the zeroes, we put $f(x)=0$ $\Rightarrow 2 x^{2}-x-6=0$ $\Rightarrow 2 x^{2}-4 x+3 x-6=0$ $\Rightarrow 2 x(x-2)+3(x-2)=0$ $\Rightarrow(x-2)(2 x+3)=0$ $\Rightarrow(x-2)=0$ or $(2 x+3)=0$ $\Rightarrow x=2,-\frac{3}{2}$ Hence, all the zeroes of the polynomial $f(x)$ are 2 and $-\frac{3}{2}$. Now, $f(2)=2(2)^{2}-2-6...
Read More →If y(x) is the solution of the differential equation
Question: If $y(x)$ is the solution of the differential equation $\frac{d y}{d x}+\left(\frac{2 x+1}{x}\right) y=e^{-2 x}, x0$, where $y(1)=\frac{1}{2} e^{-2}$, then :(1) $y\left(\log _{e} 2\right)=\log _{e} 4$(2) $y\left(\log _{e} 2\right)=\frac{\log _{e} 2}{4}$(3) $y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$(4) $\mathrm{y}(x)$ is decreasing in $(0,1)$Correct Option: , 3 Solution: Given differential equation is, $\frac{d y}{d x}+\left(2+\frac{1}{x}\right) y=e^{-2 x}, x0$ $\mathrm{IF}=e...
Read More →Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
Question: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients $x^{2}+7 x+12$ Solution: $x^{2}+7 x+12=0$ $\Rightarrow x^{2}+4 x+3 x+12=0$ $\Rightarrow x(x+4)+3(x+4)=0$ $\Rightarrow(x+4)(x+3)=0$ $\Rightarrow(x+4)=0$ or $(x+3)=0$ $\Rightarrow x=-4$ or $x=-3$ Sum of zeroes $=-4+(-3)=\frac{-7}{1}=\frac{-(\text { coefficient of } x)}{\text { (coefficient of } x^{2} \text { ) }}$ Product of zeroes $=(-4)(-3)=\frac{12}{1}=\frac{\text...
Read More →The current i in the network is:
Question: The current $i$ in the network is: (1) $0.2 \mathrm{~A}$(2) $0.6 \mathrm{~A}$(3) $0.3 \mathrm{~A}$(4) $0 \mathrm{~A}$Correct Option: , 3 Solution: (3) Both the diodes are reverse biased, so, there is no flow of current through $5 \Omega$ and $20 \Omega$ resistances. Now, two resistors of $10 \Omega$ and two resistors of $5 \Omega$ are in series. Hence current $I$ through the network $I=\frac{V}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{10+5+10+5}$ $\Rightarrow I=\frac{9}{30}=0.3 \mathrm{~A}$...
Read More →Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
Question: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients $x^{2}-2 x-8$ Solution: $x^{2}-2 x-8=0$ $\Rightarrow x^{2}-4 x+2 x-8=0$ $\Rightarrow x(x-4)+2(x-4)=0$ $\Rightarrow(x-4)(x+2)=0$ $\Rightarrow(x-4)=0$ or $(x+2)=0$ $\Rightarrow x=4$ or $x=-2$ Sum of zeroes $=4+(-2)=2=\frac{2}{1}=\frac{-(\text { coefficient of } x)}{\text { coefficient of } x^{2}}$ Product of zeroes $=4(-2)=-8=\frac{-8}{1}=\frac{\text { constant term ...
Read More →Find the zeros of the quadratic polynomial
Question: Find the zeros of the quadratic polynomial $\left(x^{2}+3 x-10\right)$ and verify the relation between its zeros and coefficients. Solution: We have: $f(x)=x^{2}+3 x-10$ $=x^{2}+5 x-2 x-10$ $=x(x+5)-2(x+5)$ $=(x-2)(x+5)$ $\therefore f(x)=0=(x-2)(x+5)=0$ $=x-2=0$ or $x+5=0$ $=x=2$ or $x=-5$ So, the zeroes of $f(x)$ are 2 and $-5$. Sum of the zeroes $=2+(-5)=-3=\frac{-3}{1}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$ Product of the zeroes $=2 \times...
Read More →Let f be a differentiable function such that
Question: Let $f$ be a differentiable function such that $f^{\prime}(x)=7-\frac{3}{4} \frac{f(x)}{x}$, $(x0)$ and $f(1) \neq 4$. Then $\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right):$(1) exists and equals $\frac{4}{7}$.(2) exists and equals 4 .(3) does not exist.(4) exists and equals $0 .$Correct Option: , 2 Solution: Let $y \quad f(x)$ $\frac{d y}{d x}+\left(\frac{3}{4 x}\right) y=7$ I.F. $=e^{\int \frac{3}{4 x} d x}=e^{\frac{3}{4} \ln x}=x^{\left(\frac{3}{4}\right)}$ Solution of diffe...
Read More →In the given circuit diagram,
Question: In the given circuit diagram, $a$ wire is joining points $B$ and D. The current in this wire is: (1) $0.4 \mathrm{~A}$(2) $2 \mathrm{~A}$(3) $4 \mathrm{~A}$(4) zeroCorrect Option: , 2 Solution: (2) From circuit diagram, $\frac{1}{R_{1}}=\frac{1}{1}+\frac{1}{4} \Rightarrow R_{1}=\frac{4}{5}$ $\frac{1}{R_{2}}=\frac{1}{2}+\frac{1}{3} \Rightarrow R_{2}=\frac{6}{5}$ $R_{\text {eff }}=R_{1}+R_{2}=\frac{4}{5}+\frac{6}{5}=2 \Omega$ $i=\frac{v}{R_{\mathrm{eff}}}=\frac{20}{2}=10 \mathrm{~A}$ $\t...
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