The distance between the points (a cos θ + b sin θ, 0)
Question: The distance between the points (a cos + b sin , 0) and (0, a sin b cos ) is (a) $a^{2}+b^{2}$ (b) $a+b$ (c) $a^{2}-b^{2}$ (d) $\sqrt{a 2+b 2}$ Solution: We have to find the distance between $\mathrm{A}(a \cos \theta+b \sin \theta, 0)$ and $\mathrm{B}(0, a \sin \theta-b \cos \theta)$. In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by, $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$...
Read More →Find the coordinates of the midpoints of the line segment joining:
Question: Find the coordinates of the midpoints of the line segment joining:(i)A(3, 0) andB(5, 4)(ii)P(11, 8) andQ(8, 2) Solution: (i) The given points areA(3, 0) andB(5, 4).Let (x,y) be the mid point ofAB. Then: $x=\frac{x_{1}+x_{2}}{2}, y=\frac{y_{1}+y_{2}}{2}$ $\Rightarrow x=\frac{3+(-5)}{2}, y=\frac{0+4}{2}$ $\Rightarrow x=\frac{-2}{2}, y=\frac{4}{2}$ $\Rightarrow x=-1, y=2$ Therefore, (1, 2) are the coordinates of mid point ofAB.(ii) The given points areP(11, 8) andQ(8, 2).Let (x,y) be the ...
Read More →Solve the following
Question: IfS1andS2denote respectively the sum of first 100 natural numbers and the sum of their cubes, then the relation betweenS1andS2is __________. Solution: Let S1: the sum of first 100 natural numbers S2: the sum of their cubes. $S_{1}=\frac{n(n+1)}{2}=\frac{100(100+1)}{2}$ $S_{1}=5050$ $S_{2}=\left(\frac{n(n+1)}{2}\right)^{2}$ $=\left(\frac{100(100+1)}{2}\right)^{2}$ $=(5050)^{2}$ $S_{2}=S_{1}^{2}$...
Read More →If x is a positive integer such that the distance between points
Question: If x is a positive integer such that the distance between points P (x, 2) and Q (3, 6) is 10 units, thenx= (a) 3(b) 3(c) 9(d) 9 Solution: It is given that distance between $P(x, 2)$ and $Q(3,-6)$ is 10 . In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by, $\mathrm{AB}^{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$ So, $10^{2}=(x-3)^{2}+(2+6)^{2}$ On further simplification, $(x-3)^{2}=36$ $x=3 \pm ...
Read More →The aperture diameter of aelescope is 5 m
Question: The aperture diameter of aelescope is $5 \mathrm{~m}$. The separation between the moon and the earth is $4 \times 10^{5} \mathrm{~km}$. With light of wavelength of $5500 A$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:(1) $60 \mathrm{~m}$(2) $20 \mathrm{~m}$(3) $200 \mathrm{~m}$(4) $600 \mathrm{~m}$Correct Option: 1 Solution: (1) Smallest angular separation between two distant objects here moon and earth, $\theta=1.22 \frac...
Read More →The distance between the points (a cos 25°, 0) and (0, a cos 65°) is
Question: The distance between the points (acos 25, 0) and (0,acos 65) is(a)a(b) 2a(c) 3a(d) None of these Solution: We have to find the distance between $\mathrm{A}\left(a \cos 25^{\circ}, 0\right)$ and $\mathrm{B}\left(0, a \cos 65^{\circ}\right)$. In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by, $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ So, $A B=\sqrt{\left(0-a \cos 25^{\circ}\rig...
Read More →Solve the following
Question: Let $S_{n}$ denote the sum of the cubes of the first $n$ natural numbers and $S_{n}$ denote the sum of the first natural numbers. Then $\sum_{r=1}^{n} \frac{S_{r}}{S_{r}}$ equals (a) $\frac{n(n+1)(n+2)}{6}$ (b) $\frac{n(n+1)}{2}$ (c) $\frac{n^{2}+3 n+2}{2}$ (d) None of these Solution: Let $S_{n}$ denote the sum of the cubes of the first $n$ natural numbers and $S_{n}$ denotes the sum of the first $n$ natural number 1 . Then $\sum_{r=1}^{n} \frac{S_{r}}{S_{r}}=?$ Since $S_{n}=\left[\fra...
Read More →The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and
Question: The line segment joining the points $A(3,-4)$ and $B(1,2)$ is trisected at the points $P(p,-2)$ and $Q\left(\frac{5}{3}, q\right)$. Find the values of $p$ and $q$. Solution: LetPandQbe the points of trisection ofAB.Then,PdividesABin the ratio 1:2.So, the coordinates ofPare $x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)}, y=\frac{\left(m y_{2}+n y_{1}\right)}{(m+n)}$ $\Rightarrow x=\frac{\{1 \times 1+2 \times(3)\}}{1+2}, y=\frac{\{1 \times 2+2 \times(-4)\}}{1+2}$ $\Rightarrow x=\frac{1+6}...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following:The distance between the points (cos , 0) and (sin cos ) is (a) $\sqrt{3}$ (b) $\sqrt{2}$ (c) 2(d) 1 Solution: We have to find the distance between $\mathrm{A}(\cos \theta, \sin \theta)$ and $\mathrm{B}(\sin \theta,-\cos \theta)$. In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by, $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ ...
Read More →The sum of the series
Question: The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\ldots . .$ to $n$ terms is (a) $n-\frac{1}{2}\left(3^{-n}-1\right)$ (b) $n-\frac{1}{2}\left(1-3^{-n}\right)$ (c) $n+\frac{1}{2}\left(3^{n}-1\right)$ (d) $n-\frac{1}{2}\left(3^{n}-1\right)$ Solution: (b) $n-\frac{1}{2}\left(1-3^{-n}\right)$ Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\frac{3^{n}-1}{3^{n}}=1-\frac{1}{3^{n}}$ Now, Let $S_{n}$ be the sum of $n$ terms of the given series...
Read More →Points P, Q and R, in that order, divide a line segment joining A(1, 6) and B(5, −2) in four equal parts.
Question: PointsP,QandR, in that order, divide a line segment joiningA(1, 6) andB(5, 2) in four equal parts. Find the coordinates ofP,QandR. Solution: The given points areA(1, 6) andB(5, 2).Then,P(x,y) is a point that divides the lineABin the ratio 1:3.By the section formula: $x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)}, y=\frac{\left(m y_{2}+n y_{1}\right)}{(m+n)}$ $\Rightarrow x=\frac{(1 \times 5+3 \times 1)}{1+3}, y=\frac{(1 \times(-2)+3 \times 6)}{1+3}$ $\Rightarrow x=\frac{5+3}{4}, y=\frac...
Read More →If A (1, 2) B (4, 3) and C (6, 6) are the three vertices
Question: IfA(1, 2)B(4, 3) andC(6, 6) are the three vertices of a parallelogramABCD, find the coordinates of fourth vertexD. Solution: Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (1, 2); B (4, 3) and C (6, 6). We have to find the co-ordinates of the forth vertex. Let the forth vertex be $\mathrm{D}(x, y)$ Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide. Now to find the mid-...
Read More →An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror.
Question: An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification $(m)$ versus distance of the object from the mirror $(x)$ is correctly given by (Graphs are drawn schematically and are not to scale)Correct Option: 3 Solution: (3) Using mirror formula, magnification is given by $m=\frac{f}{u-f}=\frac{-1}{1-\frac{u}{f}}$ At focus magnification is $\infty$ And at $u=2 f$, magnifi...
Read More →The sum of the series
Question: The sum of the series 12+ 32+ 52+ ... tonterms is (a) $\frac{n(n+1)(2 n+1)}{2}$ (b) $\frac{n(2 n-1)(2 n+1)}{3}$ (c) $\frac{(n-1)^{2}(2 n+1)}{6}$ (d) $\frac{(2 n+1)^{3}}{3}$ Solution: (b) $\frac{n(2 n-1)(2 n+1)}{3}$ Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=(2 n-1)^{2}=4 n^{2}+1-4 n$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n}\left(4 k^{2}+1-4 k\right)$ $\Rightarrow S_{n}=4 \sum_{k=1}^{n} k^{2}+\sum_{k...
Read More →If P (2, p) is the mid-point of the line segment joining the points
Question: IfP(2,p) is the mid-point of the line segment joining the pointsA(6, 5) andB(2, 11). find the value ofp. Solution: It is given that mid-point of line segment joining $A(6,-5)$ and $B(-2,11)$ is $P(2, p)$ In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as, $\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$ So, $(2, p)=\left(\frac{6-2}{2}, \fr...
Read More →If P (2, 6) is the mid-point of the line segment joining
Question: IfP(2, 6) is the mid-point of the line segment joiningA(6, 5) andB(4,y), findy. Solution: It is given that mid-point of line segment joining A (6, 5) and B (4,y) is P (2, 6) In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as, $\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$ So, $(2,6)=\left(\frac{4+6}{2}, \frac{y+5}{2}\right)$ Now equate t...
Read More →The sum of 10 terms of the series
Question: The sum of 10 terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ldots .$ is (a) $121(\sqrt{6}+\sqrt{2})$ (b) $243(\sqrt{3}+1)$ (c) $\frac{121}{\sqrt{3}-1}$ (d) $242(\sqrt{3}-1)$ Solution: (a) $121(\sqrt{6}+\sqrt{2})$ Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\sqrt{2 \times 3^{n-1}}=\sqrt{2}\left(\sqrt{3^{n-1}}\right)$ Now, let $S_{10}$ be the sum of 10 terms of the given series. Thus, we have: $S_{10}=\sqrt{2} \sum_{k=1}^{10}\left(\sqrt{3^{(k-1)}}\right)$...
Read More →Sum of n terms of the series
Question: Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots .$ is (a) $\frac{n(n+1)}{2}$ (b) $2 n(n+1)$ (c) $\frac{n(n+1)}{\sqrt{2}}$ (d) 1 Solution: (c) $\frac{n(n+1)}{\sqrt{2}}$ Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\sqrt{2 \times n^{2}}=n \sqrt{2}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sqrt{2} \sum_{k=1}^{n}(k)$ $\Rightarrow S_{n}=\sqrt{2}\left[\frac{n(n+1)}{2}\right]$ $\Rightarrow S_{...
Read More →to n terms is S, then S is equal to
Question: If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots$ to $n$ terms is $S$, then $S$ is equal to (a) $\frac{n(n+3)}{4}$ (b) $\frac{n(n+2)}{4}$ (c) $\frac{n(n+1)(n+2)}{6}$ (d) $n^{2}$ Solution: (a) $\frac{n(n+3)}{4}$ Let $T_{n}$ be the $n$th term of the given series. Thus, we have: $T_{n}=\frac{1+2+3+4+5+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n}{2}+\frac{1}{2}$ Now, let $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=\sum_{k=1}^{n}\left(\frac{k}{2}+\frac{1}{2}\right)$ $\R...
Read More →A point object in air is in front of the curved surface of a plano-convex lens.
Question: A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 \mathrm{~cm}$ and the refractive index of the lens material is $1.5$, then the focal length of the lens (in $\mathrm{cm}$ ) is Solution: (60) Given : $\mu=1.5 ; R_{\text {curved }}=30 \mathrm{~cm}$ Using, Lens-maker formula $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ For plano-convex lens $R_{1} \rightarrow \infty$ then $R_{2}=-R$ $\...
Read More →Point P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts.
Question: PointP,Q,RandSdivide the line segment joining the pointsA(1, 2) andB(6, 7) in five equal parts.Find the coordinates of the pointsP,QandR. Solution: Since, the pointsP,Q,RandSdivide the line segment joining the pointsA(1, 2) andB(6, 7) in five equal parts, soAP=PQ=QR=RS=SBHere, pointPdividesABin the ratio of 1 : 4 internally. So using section formula, we get Coordinates of $P=\left(\frac{1 \times(6)+4 \times(1)}{1+4}, \frac{1 \times(7)+4 \times(2)}{1+4}\right)$ $=\left(\frac{6+4}{5}, \f...
Read More →The magnifying power of a telescope with tube length 60 cm is 5 .
Question: The magnifying power of a telescope with tube length 60 $\mathrm{cm}$ is 5 . What is the focal length of its eye piece?$20 \mathrm{~cm}$$40 \mathrm{~cm}$$30 \mathrm{~cm}$$10 \mathrm{~cm}$Correct Option: 4 Solution: (4) For telescope Tube length $(\mathrm{L})=f_{o}+f_{e}=60$ and magnification $(m)=\frac{f_{o}}{f_{e}}=5 \Rightarrow f_{0}=5 f_{e}$ $\therefore f_{\mathrm{o}}=50 \mathrm{~cm}$ and $f_{\mathrm{e}}=10 \mathrm{~cm}$ Hence focal length of eye-piece, $f_{\mathrm{e}}=10 \mathrm{~c...
Read More →Point A lies on the line segment PQ joining P(6, −6) and Q(−4, −1) in such a way that
Question: Point $A$ lies on the line segment $P Q$ joining $P(6,-6)$ and $Q(-4,-1)$ in such a way that $\frac{P A}{P Q}=\frac{2}{5}$. If the point $A$ also lies on the line $3 x+k(y+1)=0$, find the value of $k$. Solution: Let the coordinates of $A$ be $(x, y)$. Here, $\frac{P A}{P Q}=\frac{2}{5}$. So, $P A+A Q=P Q$ $\Rightarrow P A+A Q=\frac{5 P A}{2} \quad\left[\because P A=\frac{2}{5} P Q\right]$ $\Rightarrow A Q=\frac{5 P A}{2}-P A$ $\Rightarrow \frac{A Q}{P A}=\frac{3}{2}$ $\Rightarrow \frac...
Read More →Solve the following
Question: If $S_{n}=\sum_{r=1}^{n} \frac{1+2+2^{2}+\ldots \text { Sum to } r \text { terms }}{2^{r}}$, then $S_{n}$ is equal to (a) $2^{n}-n-1$ (b) $1-\frac{1}{2^{n}}$ (c) $n-1+\frac{1}{2^{n}}$ (d) $2^{n}-1$ Solution: (c) $n-1+\frac{1}{2^{n}}$ We have: $S_{n}=\sum_{r=1}^{n} \frac{1+2+2^{2}+\ldots \text { sum to } r \text { terms }}{2^{r}}$ $\Rightarrow S_{n}=\sum_{r=1}^{n} \frac{1\left(2^{r}-1\right)}{2^{r}}$ $\Rightarrow S_{n}=\sum_{r=1}^{n}\left(1-\frac{1}{2^{r}}\right)$ $\Rightarrow S_{n}=n-\...
Read More →A thin lens made of glass (refractive index =1.5)
Question: A thin lens made of glass (refractive index $=1.5$ ) of focal length $f=16 \mathrm{~cm}$ is immersed in a liquid of refractive index 1.42. If its focal length in liquid is $f_{l}$, then the ratio $f_{l}$ /f is closest to the integer:19517Correct Option: 2 Solution: (2) Using lens maker's formula $\frac{1}{f}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$ Here, $\mu_{g}$ and $\mu_{a}$ are the refractive index of glass and air respectively ....(i) $\R...
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