1 + 4 + 13 + 40 + 121 + ...
Question: 1 + 4 + 13 + 40 + 121 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum to $n$ terms of the given series. Thus, we have: $S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n}$ ....(1) Equation (1) can be rewritten as: $S_{n}=1+4+13+40+121+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: The sequence of difference between successive terms is 3, 9, 27, 81,... We observe that it is a GP with common ratio 3 and first term 3. Thus, we have: $1+\left[\frac{3\left(3^{n...
Read More →A double convex lens has power P and same radii of curvature R of both the surfaces.
Question: A double convex lens has power $P$ and same radii of curvature $R$ of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power $1.5 P$ is :$2 R$$\frac{R}{2}$$\frac{3 R}{2}$$\frac{R}{3}$Correct Option: 4 Solution: (4) Given, using lens maker's formula $\frac{1}{f}=(k-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ Here, $R_{1}=R_{2}=R$ (For double convex lens) $\therefore \frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$...
Read More →Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram.
Question: Show thatA(1, 2),B(4, 3),C(6, 6) andD(3, 5) are the vertices of a parallelogram. Show thatABCDis not a rectangle. Solution: The given vertices areA(1, 2),B(4, 3),C(6, 6) andD(3, 5). $A B=\sqrt{(1-4)^{2}+(2-3)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}$ $=\sqrt{9+1}=\sqrt{10}$ $B C=\sqrt{(4-6)^{2}+(3-6)^{2}}=\sqrt{(-2)^{2}+(-3)^{2}}$ $=\sqrt{4+9}=\sqrt{13}$ $C D=\sqrt{(6-3)^{2}+(6-5)^{2}}=\sqrt{(3)^{2}+(1)^{2}}$ $=\sqrt{9+1}=\sqrt{10}$ $A D=\sqrt{(1-3)^{2}+(2-5)^{2}}=\sqrt{(-2)^{2}+(-3)^{2}}$ $=\sqr...
Read More →Write the coordinates of a point on X-axis which is equidistant
Question: Write the coordinates of a point on X-axis which is equidistant from the points (3, 4) and (2, 5). Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ Here we are to find out a point on thexaxis which is equidistant from both the points A(-3,4) and B(2,5). Let this point be denoted as C(x, y). Since the point lies on thex-axis the value of i...
Read More →1 + 3 + 6 + 10 + 15 + ...
Question: 1 + 3 + 6 + 10 + 15 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=1+3+6+10+15+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=1+3+6+10+15+\ldots+T_{n-1}+T_{n}$ ....(2) On subtracting (2) from (1), we get: The sequence of difference of successive terms is 2, 3, 4, 5,... We observe that it is an AP with common difference 1 and first term 2. Thus, we have: $1+\left[\frac{(n-1)}{2}(4+(n-2) ...
Read More →A point like object is placed at a distance of 1 m
Question: A point like object is placed at a distance of $1 \mathrm{~m}$ in front of a convex lens of focal length $0.5 \mathrm{~m}$. A plane mirror is placed at a distance of $2 \mathrm{~m}$ behind the lens. The position and nature of the final image formed by the system is :$2.6 \mathrm{~m}$ from the mirror, real$1 \mathrm{~m}$ from the mirror, virtual$1 \mathrm{~m}$ from the mirror, real$2.6 \mathrm{~m}$ from the mirror, virtualCorrect Option: 4 Solution: (4) Focal length of the convex lens, ...
Read More →Solve the following
Question: In Q. No. 9, what is the value of $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ ? Solution: The co-ordinates of the vertices are (a, b); (b, c) and (c, a) The co-ordinate of the centroid is (0, 0) We know that the co-ordinates of the centroid of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is- $\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$ So, $(0,0)=\left(\frac{a+b+c}{3}, \frac{b+c+a}{3...
Read More →Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram.
Question: Show that the pointsA(2, 1),B(5, 2),C(6, 4) andD(3, 3) are the angular points of a parallelogram. Is this figure a rectangle? Solution: The given points areA(2, 1),B(5, 2),C(6, 4) andD(3, 3). $A B=\sqrt{(5-2)^{2}+(2-1)^{2}}=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10}$ units $B C=\sqrt{(6-5)^{2}+(4-2)^{2}}=\sqrt{(1)^{2}+(2)^{2}}=\sqrt{1+4}=\sqrt{5}$ units $C D=\sqrt{(3-6)^{2}+(3-4)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9+1}=\sqrt{10}$ units $A D=\sqrt{(3-2)^{2}+(3-1)^{2}}=\sqrt{(1)^{2}+(2...
Read More →3 + 7 + 14 + 24 + 37 + ...
Question: 3 + 7 + 14 + 24 + 37 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=3+7+14+24+37+\ldots+T_{n-1}+T_{n} \quad \ldots(1)$ Equation (1) can be rewritten as: $S_{n}=3+7+14+24+37+\ldots+T_{n-1}+T_{n} \quad \ldots$ (2) On subtracting (2) from (1), we get: The sequence of difference of successive terms is 4, 7, 10, 13,... We observe that it is an AP with common difference 3 and first term 4. Thus, we have: $3+\left[\...
Read More →3 + 7 + 14 + 24 + 37 + ...
Question: 3 + 7 + 14 + 24 + 37 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=3+7+14+24+37+\ldots+T_{n-1}+T_{n} \quad \ldots(1)$ Equation (1) can be rewritten as: $S_{n}=3+7+14+24+37+\ldots+T_{n-1}+T_{n} \quad \ldots$ (2) On subtracting (2) from (1), we get: The sequence of difference of successive terms is 4, 7, 10, 13,... We observe that it is an AP with common difference 3 and first term 4. Thus, we have: $3+\left[\...
Read More →If the centroid of the triangle formed by points
Question: If the centroid of the triangle formed by points P (a, b), Q(b, c) and R (c, a) is at the origin, what is the value of a + b + c? Solution: The co-ordinates of the vertices are (a, b); (b, c) and (c, a) The co-ordinate of the centroid is (0, 0) We know that the co-ordinates of the centroid of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$ is- $\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$ So, $(0,0)...
Read More →Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus.
Question: Show that the pointsA(6, 1),B(8, 2),C(9, 4) andD(7, 3) are the vertices of a rhombus. Find its area. Solution: The given points areA(6, 1),B(8, 2),C(9, 4) andD(7, 3). $A B=\sqrt{(6-8)^{2}+(1-2)^{2}}=\sqrt{(-2)^{2}+(-1)^{2}}$ $=\sqrt{4+1}=\sqrt{5}$ $B C=\sqrt{(8-9)^{2}+(2-4)^{2}}=\sqrt{(-1)^{2}+(-2)^{2}}$ $=\sqrt{1+4}=\sqrt{5}$ $C D=\sqrt{(9-7)^{2}+(4-3)^{2}}=\sqrt{(2)^{2}+(1)^{2}}$ $=\sqrt{4+1}=\sqrt{5}$ $A D=\sqrt{(7-6)^{2}+(3-1)^{2}}=\sqrt{(1)^{2}+(2)^{2}}$ $=\sqrt{1+4}=\sqrt{5}$ $A ...
Read More →A prism of angle
Question: A prism of angle $\mathrm{A}=1^{\circ}$ has a refractive index $\mu=1.5$. A good estimate for the minimum angle of deviation (in degrees) is close to $N / 10$. Value of $N$ is___________ Solution: (5) Given, Angle of prism, $A=1^{\circ}$ Refractive index of prism, $\mu=1.5$ For a thin prism, minimum angle of deviation is given by $\delta=(\mu-1) A$ $\Rightarrow \delta=(1.5-1) \times 1^{\circ}=\frac{1}{2}=\frac{5}{10}$ $\Rightarrow N=5$...
Read More →Write the coordinates of the point dividing line segment
Question: Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5. Solution: Let $P(x, y)$ be the point which divide the line segment joining $A(2,3)$ and $B(3,4)$ in the ratio $1: 5$. Now according to the section formula if point a point $\mathrm{P}$ divides a line segment joining $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ in the ratio $\mathrm{m}$ : $\mathrm{n}$ internally than, $\mathrm{P}(x, ...
Read More →What is the area of the triangle formed by
Question: What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4)? Solution: The given triangle $\Delta \mathrm{OAB}$ is a right angled triangle, right angled at $O$. the co-ordinates of the vertices are $O(0,0) A(6,0)$ and $B(0,4)$. So, Altitude is 6 units and base is 4 units. Therefore, $\operatorname{ar}(\Delta \mathrm{OAB})=\frac{1}{2}($ Base $)($ Altitude $)$ $=\frac{1}{2}(4)(6)$ sq. units $=12$ sq. units...
Read More →3 + 5 + 9 + 15 + 23 + ...
Question: 3 + 5 + 9 + 15 + 23 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum to $n$ terms of the given series. Thus, we have: $S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=3+5+9+15+23+\ldots+T_{n-1}+T_{n} \ldots(2)$ On subtracting (2) from (1), we get: The sequence of difference of successive terms is 2, 4, 6, 8,... We observe that it is an AP with common difference 2 and first term 2. Thus, we have: $3+\left[\frac{(n-1)}{2}\{4+(n-...
Read More →A compound microscope consists of an objective lens of focal length 1 cm
Question: A compound microscope consists of an objective lens of focal length $1 \mathrm{~cm}$ and an eye piece of focal length $5 \mathrm{~cm}$ with a separation of $10 \mathrm{~cm}$. The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{n}{40} \mathrm{~cm}$. The value of $n$ is___________ Solution: (50) Given : Length of compound microscope, $L=10 \mathrm{~cm}$ Focal length of objective $f_{0}=1 \mathrm{~cm}$ and of eye-piece, $f_{e}=5$ $\ma...
Read More →Show that the points A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1) are the vertices of a rhombus.
Question: Show that the pointsA(3, 0),B(4, 5),C(1, 4) andD(2, 1) are the vertices of a rhombus. Find its area. Solution: The given points areA(3, 0),B(4, 5),C( 1, 4) andD( 2, 1). $A B=\sqrt{(3-4)^{2}+(0-5)^{2}}=\sqrt{(-1)^{2}+(-5)^{2}}$ $=\sqrt{1+25}=\sqrt{26}$ $B C=\sqrt{(4+1)^{2}+(5-4)^{2}}=\sqrt{(5)^{2}+(1)^{2}}$ $=\sqrt{25+1}=\sqrt{26}$ $C D=\sqrt{(-1+2)^{2}+(4+1)^{2}}=\sqrt{(1)^{2}+(5)^{2}}$ $=\sqrt{1+25}=\sqrt{26}$ $A D=\sqrt{(3+2)^{2}+(0+1)^{2}}=\sqrt{(5)^{2}+(1)^{2}}$ $=\sqrt{25+1}=\sqrt...
Read More →2 + 5 + 10 + 17 + 26 + ...
Question: 2 + 5 + 10 + 17 + 26 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=2+5+10+17+26+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: The sequence of difference of successive terms is 3, 5, 7, 9,...We observe that it is an AP with common difference 2 and first term 3. Thus, we have: $2+\left[\frac{(n-1)}{2}\{6+(n-2...
Read More →If the distance between points (x, 0)
Question: If the distance between points (x, 0) and (0, 3) is 5, what are the values of x? Solution: We have to find the unknown $x$ using the distance between $\mathrm{A}(x, 0)$ and $\mathrm{B}(0,3)$ which is $5 .$ 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, $5=\sqrt{(x-0)^{2}+(0-3)^{2}}$ Squaring both the sides we get, $x^{2}-16=0$...
Read More →1 + 3 + 7 + 13 + 21 + ...
Question: 1 + 3 + 7 + 13 + 21 + ... Solution: Let $T_{n}$ be the $n$th term and $S_{n}$ be the sum of $n$ terms of the given series. Thus, we have: $S_{n}=1+3+7+13+21+\ldots+T_{n-1}+T_{n}$ ...(1) Equation (1) can be rewritten as: $S_{n}=1+3+7+13+21+\ldots+T_{n-1}+T_{n}$ ...(2) On subtracting (2) from (1), we get: The sequence of difference of successive terms is 2, 4, 6, 8,...We observe that it is an AP with common difference 2 and first term 2. Thus, we have: $1+\left[\frac{(n-1)}{2}\{4+(n-2) 2...
Read More →For a concave lens of focal length f,
Question: For a concave lens of focal length $f$, the relation between object and image distances $u$ and $v$, respectively, from its pole can best be represented by ( $u=v$ is the reference line):Correct Option: 4 Solution: (4) From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow v=\frac{u f}{u+f}$ Case-I : If $v=u \Rightarrow f+u=f \Rightarrow u=0$ Case-II : If $u=\infty$ then $v=f$. Hence, correct $u$ versus $v$ graph, that satisfies this condition is (a)....
Read More →If A (−1, 3) , B(1, −1) and C (5, 1) are the vertices of a triangle ABC,
Question: IfA(1, 3) ,B(1, 1) andC(5, 1) are the vertices of a triangleABC, what is the length of the median through vertexA? Solution: We have a triangle $\triangle \mathrm{ABC}$ in which the co-ordinates of the vertices are $\mathrm{A}(-1,3) \mathrm{B}(1,-1)$ and $\mathrm{C}(5,1)$. 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}}{...
Read More →Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus.
Question: Show that the pointsA(3, 2),B(5, 5),C(2, 3) andD(4, 4) are the vertices of a rhombus. Find the area of this rhombus. Solution: The given points areA(3, 2),B(5, 5),C(2, 3) andD(4, 4). $A B=\sqrt{(-5+3)^{2}+(-5-2)^{2}}=\sqrt{(-2)^{2}+(-7)^{2}}=\sqrt{4+49}=\sqrt{53}$ units. $B C=\sqrt{(2+5)^{2}+(-3+5)^{2}}=\sqrt{(7)^{2}+(2)^{2}}=\sqrt{49+4}=\sqrt{53}$ units. $C D=\sqrt{(4-2)^{2}+(4+3)^{2}}=\sqrt{(2)^{2}+(7)^{2}}=\sqrt{4+49}=\sqrt{53}$ units. $D A=\sqrt{(4+3)^{2}+(4-2)^{2}}=\sqrt{(7)^{2}+(...
Read More →What is the distance between the points
Question: What is the distance between the points (5 sin 60, 0) and (0, 5 sin 30)? Solution: We have to find the distance between $A\left(5 \sin 60^{\circ}, 0\right)$ and $B\left(0,5 \sin 30^{\circ}\right)$. 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, $\mathrm{AB}=\sqrt{\left(5 \sin 60^{\circ}-0\right)^{2}+\left(0-5 \sin 30^{\circ}\r...
Read More →