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Question: $\frac{\cos \theta \operatorname{cosec} \theta-\sin \theta \sec \theta}{\cos \theta+\sin \theta}=\operatorname{cosec} \theta-\sec \theta$ Solution: $\mathrm{LHS}=\frac{\cos \theta \operatorname{cosec} \theta-\sin \theta \sec \theta}{\cos \theta+\sin \theta}$ $=\frac{\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}}{\cos \theta+\sin \theta}$ $=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta \sin \theta(\cos \theta+\sin \theta)}$ $=\frac{(\cos \theta+\sin \theta)(\cos...
Read More →A boy is moving on a straight road against
Question: A boy is moving on a straight road against a frictional force of $5 \mathrm{~N}$. After travelling a distance of $1.5 \mathrm{~km}$ he forgot the correct path at a round about (Fig. 1) of radius $100 \mathrm{~m}$. Solution: However, he moves on the circular path for one and half cycle and then he moves forward upto $2.0 \mathrm{~km}$. Calculate the work done by him. Here, $F=5 \mathrm{~N}$, Distance travelled $S=1500+3 \pi r+2000=4442.86 m$ $\therefore W=F \times S=5 \times 4442.86=222...
Read More →Find the square root of:
Question: Find the square root of: (i) $\frac{441}{961}$ (ii) $\frac{324}{841}$ (iii) $4 \frac{29}{29}$ (iv) $2 \frac{14}{25}$ (v) $2 \frac{137}{196}$ (vi) $23 \frac{26}{121}$ (vii) $25 \frac{544}{729}$ (viii) $75 \frac{46}{49}$ (ix) $3 \frac{942}{2209}$ (X) $3 \frac{334}{3025}$ (xi) $21 \frac{2797}{3364}$ (xii) $38 \frac{11}{25}$ (xiii) $23 \frac{394}{729}$ (xiv) $21 \frac{51}{169}$ (XV) $10 \frac{151}{225}$ Solution: (i) We know: $\sqrt{\frac{441}{961}}=\frac{\sqrt{441}}{\sqrt{961}}$ Now, let ...
Read More →Avinash can run with a speed of
Question: Avinash can run with a speed of $8 \mathrm{~m} \mathrm{~s}^{-1}$ against the frictional force of $10 \mathrm{~N}$, and Kapil can move with a speed of $3 \mathrm{~ms}^{-1}$ against the frictional force of $25 \mathrm{~N}$. Who is more powerful and why? Solution: $P=F u$ Power of Avinash $=10 \times 8=80 \mathrm{~W}$ Power of Kapil $=25 \times 3=75 \mathrm{~W}$ So, Avinash is more powerful....
Read More →Find the square root of:
Question: Find the square root of: (i) $\frac{441}{961}$ (ii) $\frac{324}{841}$ (iii) $4 \frac{29}{29}$ (iv) $2 \frac{14}{25}$ (v) $2 \frac{137}{196}$ (vi) $23 \frac{26}{121}$ (vii) $25 \frac{544}{729}$ (viii) $75 \frac{46}{49}$ (ix) $3 \frac{942}{2209}$ (X) $3 \frac{334}{3025}$ (xi) $21 \frac{2797}{3364}$ (xii) $38 \frac{11}{25}$ (xiii) $23 \frac{394}{729}$ (xiv) $21 \frac{51}{169}$ (XV) $10 \frac{151}{225}$ Solution: (i) We know: $\sqrt{\frac{441}{961}}=\frac{\sqrt{441}}{\sqrt{961}}$ Now, let ...
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Question: If $A=\left[\begin{array}{ll}1 2 \\ 0 3\end{array}\right]$ is written as $B+C$, where $B$ is a symmetric matrix and $C$ is a skew-symmetric matrix, then $B$ is equal to. Solution: Given : $A=\left[\begin{array}{ll}1 2 \\ 0 3\end{array}\right]$ $\Rightarrow A^{T}=\left[\begin{array}{ll}1 0 \\ 2 3\end{array}\right]$ Let $B=\frac{1}{2}\left(A+A^{T}\right)=\frac{1}{2}\left(\left[\begin{array}{ll}1 2 \\ 0 3\end{array}\right]+\left[\begin{array}{ll}1 0 \\ 2 3\end{array}\right]\right)$ $=\fra...
Read More →A rocket is moving up with a velocity u.
Question: A rocket is moving up with a velocity $u$. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies? Solution: $\mathrm{K}_{1}=\frac{1}{2} m v^{2}$ $\mathrm{K}_{2}=\frac{1}{2} m(3 v)^{2}=\frac{9}{2} m v^{2}$ $\therefore \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{1}{9}$ or $\mathrm{K}_{2}=9 \mathrm{~K}_{1}$...
Read More →Prove that
Question: $\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta$ Solution: $\mathrm{LHS}=\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}$ $=\frac{(1+\cos \theta)-\left(1-\cos ^{2} \theta\right)}{\sin \theta(1+\cos \theta)}$ $=\frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta)}$ $=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$ $=\frac{\cos \theta}{\sin \theta}$ $=\cot \theta$ $=\mathrm{RHS}$ Hence, L.H.S. = R.H.S....
Read More →A body is falling from a height h.
Question: A body is falling from a height $h$. After it has fallen a height h/2, it will possess (a) only potential energy (b) only kinetic energy (c) half potential and half kinetic energy (d) more kinetic and less potential energy. Solution: (c) half potential and half kinetic energy...
Read More →Water stored in a dam possesses
Question: Water stored in a dam possesses (a) no energy (b) electrical energy (c) kinetic energy (d) potential energy Solution: (d). The energy possessed by a body by virtue of its position is called potential energy....
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Question: $\frac{\cos ^{3} \theta+\sin ^{3} \theta}{\cos \theta+\sin \theta}+\frac{\cos ^{3} \theta-\sin ^{3} \theta}{\cos \theta-\sin \theta}=2$ Solution: $\mathrm{LHS}=\frac{\cos ^{3} \theta+\sin ^{3} \theta}{\cos \theta+\sin \theta}+\frac{\cos ^{3} \theta-\sin ^{3} \theta}{\cos \theta-\sin \theta}$ $=\frac{(\cos \theta+\sin \theta)\left(\cos ^{2} \theta-\cos \theta \sin \theta+\sin ^{2} \theta\right)}{(\cos \theta+\sin \theta)}+\frac{(\cos \theta-\sin \theta)\left(\cos ^{2} \theta+\cos \theta...
Read More →The work done on an object does not depend upon the
Question: The work done on an object does not depend upon the (a) displacement (b) force applied (c) angle between force and displacement (d) initial velocity of the object. Solution: (d) $W=F S \cos \theta$....
Read More →Which one of the following is not the unit of energy?
Question: Which one of the following is not the unit of energy? (a) joule (b) newton metre (c) kiiowatt (d) kolwatt hour. Solution: (c) It is the unit of power....
Read More →A girl is carrying a school bag of
Question: A girl is carrying a school bag of $3 \mathrm{~kg}$ mass on her back and moves $200 \mathrm{~m}$ on a levelled road. The work done against the gravitational force will be $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$ (a) $6 \times 10^{3} \mathrm{~J}$ (b) $6 \mathrm{~J}$ (c) $0.6 \mathrm{~J}$ (d) zero. Solution: (a). Freely falling bodies moves with constant acceleration....
Read More →An iron sphere of mass 10kg has the same
Question: An iron sphere of mass $10 \mathrm{~kg}$ has the same diameter as an aluminium sphere of mass is $3.5 \mathrm{~kg}$. Both spheres are dropped simultaneously from a tower. When they are $10 \mathrm{~m}$ above the ground, they have the same (a) acceleration (b) momenta (c) potential energy (d) kinetic energy. Solution: (a). Freely falling bodies moves with constant acceleration....
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Question: $\frac{(\sec \theta-\tan \theta)}{(\sec \theta+\tan \theta)}=\left(1+2 \tan ^{2} \theta-2 \sec \theta \tan \theta\right)$ Solution: $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}$ $=\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$ $=\frac{(\sec \theta-\tan \theta)^{2}}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$ $=\frac{\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}{\sec ^{2} \theta-...
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Question: $\frac{(\sec \theta-\tan \theta)}{(\sec \theta+\tan \theta)}=\left(1+2 \tan ^{2} \theta-2 \sec \theta \tan \theta\right)$ Solution: $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}$ $=\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$ $=\frac{(\sec \theta-\tan \theta)^{2}}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$ $=\frac{\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}{\sec ^{2} \theta-...
Read More →In case of negative work,
Question: In case of negative work, the angle between the force and displacement is (a) $0^{\circ}$ (b) $45^{\circ}$ (c) $90^{\circ}$ (d) $180^{\circ}$. Solution: (d) Explanation: $W=F S \cos \theta .$ When $\theta=180^{\circ}, \cos 180^{\circ}=-1$ and $W=-F S$...
Read More →A car is accelerated on a levelled road and attains a velocity 4 times
Question: A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process, the potential energy of the car (a) does not change (b) becomes twice to that of initial (c) becomes 4 times that of initial (d) becomes 16 times that of initial. Solution: (a). Potential energy does not depend on the velocity of a body....
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Question: $\frac{1+\cos \theta}{1-\cos \theta}=(\operatorname{cosec} \theta+\cot \theta)^{2}$ Solution: $\frac{1+\cos \theta}{1-\cos \theta}$ $=\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}$ $=\frac{(1+\cos \theta)^{2}}{(1-\cos \theta)(1+\cos \theta)}$ $=\frac{(1+\cos \theta)^{2}}{1-\cos ^{2} \theta} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$ $=\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta} \quad\left(\sin ^{2} \theta+\cos ^{2} \theta=1\right)$ $=\left(\frac{1+\cos ...
Read More →When a body falls freely towards the earth,
Question: When a body falls freely towards the earth, then its total energy (a) increases (b) decreases (c) remains constant (d) first increases and then decreases, Solution: (c) Total energy = K.E. + P.E. When a body falls freely, its K.E. increases and P.E. decreases but the sum of K.E and P.E. remains the same....
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Question: $\frac{1-\sin \theta}{1+\sin \theta}=(\sec \theta-\tan \theta)^{2}$ Solution: $\frac{1-\sin \theta}{1+\sin \theta}$ $=\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$ $=\frac{(1-\sin \theta)^{2}}{(1+\sin \theta)(1-\sin \theta)}$ $=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$ $=\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta} \quad\left(\sin ^{2} \theta+\cos ^{2} \theta=1\right)$ $=\left(\frac{1-\sin \theta}{\cos \th...
Read More →Prove that
Question: $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$ Solution: $\frac{\cot A-\cos A}{\cot A+\cos A}$ $=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}$ $=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}$ $=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} \quad\left(\operatorname{cosec} A=\frac{1}{\sin A}\right)$...
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Question: $\frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}$ Solution: $\frac{\tan A+\sin A}{\tan A-\sin A}$ $=\frac{\frac{\sin A}{\cos A}+\sin A}{\frac{\sin A}{\cos A}-\sin A}$ $=\frac{\sin A\left(\frac{1}{\cos A}+1\right)}{\sin A\left(\frac{1}{\cos A}-1\right)}$ $=\frac{\sec A+1}{\sec A-1} \quad\left(\sec A=\frac{1}{\cos A}\right)$...
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Question: $\sec ^{2} \theta-\frac{\sin ^{2} \theta-2 \sin ^{4} \theta}{2 \cos ^{4} \theta-\cos ^{2} \theta}=1$ Solution: $\sec ^{2} \theta-\frac{\sin ^{2} \theta-2 \sin ^{4} \theta}{2 \cos ^{4} \theta-\cos ^{2} \theta}$ $=\sec ^{2} \theta-\frac{\sin ^{2} \theta\left(1-2 \sin ^{2} \theta\right)}{\cos ^{2} \theta\left(2 \cos ^{2} \theta-1\right)}$ $=\sec ^{2} \theta-\frac{\sin ^{2} \theta\left(1-2 \sin ^{2} \theta\right)}{\cos ^{2} \theta\left[2\left(1-\sin ^{2} \theta\right)-1\right]} \quad\left[...
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