An object is dropped from rest at the height of
Question: An object is dropped from rest at the height of $150 \mathrm{~m}$ and simultaneously another object is dropped from rest at the height of $100 \mathrm{~m}$. What is the difference in their heights after $2 \mathrm{~s}$ if both the objects drop with same accelerations? How does the difference in heights vary with time? Solution: Distance moved by the object dropped from height of $150 \mathrm{~m}$ in 2 seconds can be calculated using $\mathrm{S}=u t+\frac{1}{2} a t^{2}$ Here, $u=0, a=g=...
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Question: If $\left[\begin{array}{cc}x 3 x-y \\ 2 x+z 3 y-\omega\end{array}\right]=\left[\begin{array}{ll}3 2 \\ 4 7\end{array}\right]$, find $x, y, z, \omega$. Solution: Since all the corresponding elements of a matrix are equal, $\left[\begin{array}{cc}x 3 x-y \\ 2 x+z 3 y-w\end{array}\right]=\left[\begin{array}{ll}3 2 \\ 4 7\end{array}\right]$ $x=3$ .....(1) $3 x-y=2$ ....(2) Putting the value of $x$ in eq. (2), we get $3(3)-y=2$ $\Rightarrow 9-y=2$ $\Rightarrow-y=-7$ $\Rightarrow y=7$ $2 x+z...
Read More →Find the squares of the following numbers:
Question: Find the squares of the following numbers: (i) 425 (ii) 575 (iii) 405 (iv) 205 (v) 95 (vi) 745 (vii) 512 (viii) 995 Solution: Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the formn5 is a number ending with 25 and has the numbern(n+ 1) before 25. (i) Here,n= 42 $\therefore n(n+1)=(42)(43)=1806$ $\therefore 425^{2}=180625$ (ii) Here,n= 57 $\therefore n(n+1)=(57)(58)=3306$ $\therefore 575^{2}=33062...
Read More →Draw a velocity versus time graph of a stone
Question: Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height Solution: When stone is thrown vertically upwards, it has some initial velocity (u). The velocity of the stone goes on decreasing as it goes upwards and becomes zero at the maximum height. There after, stone begins to fall and its velocity goes on increasing but in opposite direction and becomes equal to the initial velocity (u) when it reaches the point o...
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Question: $\frac{\cos 12^{\circ}}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}}=?$ (a) 0(b) 1(c) 2 (d) $\frac{3}{2}$ Solution: $\frac{\cos 12^{\circ}}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}}=\frac{\cos \left(90^{\circ}-78^{\circ}\right)}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}}$ $=\frac{\sin 78^{\circ}}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$ $=1+\frac{\tan \le...
Read More →Find the squares of the following numbers:
Question: Find the squares of the following numbers: (i) 425 (ii) 575 (iii) 405 (iv) 205 (v) 95 (vi) 745 (vii) 512 (viii) 995 Solution: Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the formn5 is a number ending with 25 and has the numbern(n+ 1) before 25. (i) Here,n= 42 $\therefore n(n+1)=(42)(43)=1806$ $\therefore 425^{2}=180625$ (ii) Here,n= 57 $\therefore n(n+1)=(57)(58)=3306$ $\therefore 575^{2}=33062...
Read More →The velocity-time graph (Fig. 6 ) shows the motion of a cyclist.
Question: The velocity-time graph (Fig. 6 ) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds. Solution: (i) Acceleration $=\frac{\text { Change in velocity }}{\text { time }}=0$ (ii) Velocity $=20 \mathrm{~m} \mathrm{~s}^{-1}$ (iii) Distance $=$ area under $v-t$ graph $=v \times t=20 \mathrm{~ms}^{-1} \times 15 \mathrm{~s}=300 \mathrm{~m} .$...
Read More →Find x, y and z so that A = B, where
Question: Find $x, y$ and $z$ so that $A=B$, where $A=\left[\begin{array}{ccc}x-2 3 2 z \\ 18 z y+2 6 z\end{array}\right], B=\left[\begin{array}{ccc}y z 6 \\ 6 y x 2 y\end{array}\right]$ Solution: Since all the corresponding elements of a matrix are equal, $A=\left[\begin{array}{ccc}x-2 3 2 z \\ 18 z y+2 6 z\end{array}\right], B=\left[\begin{array}{ccc}y z 6 \\ 6 y x 2 y\end{array}\right]$ Here, $x-2=y \quad \ldots(1)$ $z=3 \quad \ldots(2)$ $18 z=6 y \quad \ldots(3)$ Putting the value of $z$ in ...
Read More →A motorcyclist drives from A to B,
Question: A motorcyclist drives from $A$ to $B$ with a uniform speed of $30 \mathrm{~km} \mathrm{~h}^{-1}$ and returns back with a speed of $20 \mathrm{~km}$ $\mathrm{h}^{-1}$. Find its average speed. Solution: Let, distance between $\mathrm{A}$ and $\mathrm{B}=x$. Time taken to go from $\mathrm{A}$ to $\mathrm{B}, t_{1}=\frac{x}{30} h$ Time taken to go from $\mathrm{B}$ to $\mathrm{A}, t_{2}=\frac{x}{20} b$ $\therefore v_{a v}=\frac{\text { Total distance }}{\text { Total time }}=\frac{2 x}{t_{...
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Question: $\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}=?$ (a) $\frac{2}{3}$ (b) $\frac{16}{29}$ (c) $\frac{16}{3}$ (d) 1 Solution: $\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}=\frac{\sec \left(90^{\circ}-58^{\circ}\right)}{\operatorname{cosec} 58^{\circ}}$ $=\frac{\operatorname{cosec} 58^{\circ}}{\operatorname{cosec} 58^{\circ}} \quad\left(\because \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta\right)$ $=1$ Hence, the correct option is (d)....
Read More →Find the squares of the following numbers:
Question: Find the squares of the following numbers: (i) 127 (ii) 503 (iii) 451 (iv) 862 (v) 265 Solution: We will use visual method as it is the most efficient method to solve this problem. (i) We have: 127 = 120 + 7 Hence, let us draw a square having side 127 units. Let us split it into 120 units and 7 units. Hence, the square of 127 is 16129. (ii) We have: 503 = 500 + 3 Hence, let us draw a square having side 503 units. Let us split it into 500 units and 3 units. Hence, the square of 503 is 2...
Read More →Find x, y and z so that A = B, where
Question: Find $x, y$ and $z$ so that $A=B$, where $A=\left[\begin{array}{ccc}x-2 3 2 z \\ 18 z y+2 6 z\end{array}\right], B=\left[\begin{array}{ccc}y z 6 \\ 6 y x 2 y\end{array}\right]$ Solution: Since all the corresponding elements of a matrix are equal, $A=\left[\begin{array}{ccc}x-2 3 2 z \\ 18 z y+2 6 z\end{array}\right], B=\left[\begin{array}{ccc}y z 6 \\ 6 y x 2 y\end{array}\right]$ Here, $x-2=y \quad \ldots(1)$ $z=3 \quad \ldots(2)$ $18 z=6 y \quad \ldots(3)$ Putting the value of $z$ in ...
Read More →sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°
Question: sin (50 + ) cos (40 ) + tan 1 tan 10 tan 20 tan 70 tan 80 tan 89 Solution: $\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$ $=\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)+\tan 1^{\circ} \tan 10^{\circ} \tan 20^{\circ} \tan \left(90^{\circ}-20^{\circ}\right) \tan \left(90^{\circ}-10^{\circ}\right) \tan \left(90^{\circ}-1^{\circ}\right)$ $=\si...
Read More →Find the squares of the following numbers using diagonal method:
Question: Find the squares of the following numbers using diagonal method: (i) 98 (ii) 273 (iii) 348 (iv) 295 (v) 171 Solution: (i) $\therefore 98^{2}=9604$ (ii) $\therefore 273^{2}=74529$ (iii) $\therefore 348^{2}=121104$ (iv) $\therefore 295^{2}=87025$ (v) $\therefore 171^{2}=29241$...
Read More →A car starts from rest and moves along the x-axis
Question: A car starts from rest and moves along the x-axis with constant acceleration $5 \mathrm{~m} \mathrm{~s}^{-2}$ for $8 \mathrm{~seconds}$. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest ? Solution: Here, $u=0, a=5 \mathrm{~m} \mathrm{~s}^{-2}, t=8 \mathrm{~s} .$ Distance travelled in first $8 \mathrm{~s}, x_{1}=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times 5 \times 64=160 \mathrm{~m} .$ Velocity after $8 \mathrm{~...
Read More →Find the values of a, b, c and d from the following equations:
Question: Find the values ofa,b,canddfrom the following equations: $\left[\begin{array}{cc}2 a+b a-2 b \\ 5 c-d 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 -3 \\ 11 24\end{array}\right]$ Solution: Since all the corresponding elements of a matrix are equal, $\left[\begin{array}{cc}2 a+b a-2 b \\ 5 c-d 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 -3 \\ 11 24\end{array}\right]$ $\Rightarrow 2 a+b=4$ $\Rightarrow b=4-2 a$ ....(1) $a-2 b=-3$ ....(2) Putting the value of $b$ in eq. (2), w...
Read More →Find the values of a, b, c and d from the following equations:
Question: Find the values ofa,b,canddfrom the following equations: $\left[\begin{array}{cc}2 a+b a-2 b \\ 5 c-d 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 -3 \\ 11 24\end{array}\right]$ Solution: Since all the corresponding elements of a matrix are equal, $\left[\begin{array}{cc}2 a+b a-2 b \\ 5 c-d 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 -3 \\ 11 24\end{array}\right]$ $\Rightarrow 2 a+b=4$ $\Rightarrow b=4-2 a$ ....(1) $a-2 b=-3$ ....(2) Putting the value of $b$ in eq. (2), w...
Read More →A girl walks along a straight path to drop
Question: A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in Fig. 4. Plot a velocity-time graph for the same. Solution: For 0 to $50 \mathrm{~s}, \mathrm{v}=\frac{100}{50}=2 \mathrm{~m} \mathrm{~s}^{-1}$ For $50 s$ to $100 s, v=\frac{-100}{50}=-2 \mathrm{~m} \mathrm{~s}^{-1}$. Velocity-time graph is shown in figure 5 ....
Read More →Find the values of a, b, c and d from the following equations:
Question: Find the values ofa,b,canddfrom the following equations: $\left[\begin{array}{cc}2 a+b a-2 b \\ 5 c-d 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 -3 \\ 11 24\end{array}\right]$ Solution: Since all the corresponding elements of a matrix are equal, $\left[\begin{array}{cc}2 a+b a-2 b \\ 5 c-d 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 -3 \\ 11 24\end{array}\right]$ $\Rightarrow 2 a+b=4$ $\Rightarrow b=4-2 a$ ....(1) $a-2 b=-3$ ....(2) Putting the value of $b$ in eq. (2), w...
Read More →How will the equations of motion for an object
Question: How will the equations of motion for an object moving with a uniform velocity change? Solution: Equations of motion of a uniformly accelerated motion of an object are 1. $V=u+a t$, 2. $S=u t+1 / 2 a t^{2}$, 3. $v^{2}-u^{2}=2 a S$. When object moves with a uniform velocity, its acceleration, $a=0$. Hence equations of motion become 1. $V=U$, 2. $S=u t$, 3. $v^{2}=u^{2}$....
Read More →Find the squares of the following numbers using column method.
Question: Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication: (i) 25 (ii) 37 (iii) 54 (iv) 71 (v) 96 Solution: (i) Here,a= 2,b= 5 Step 1. Make 3 columns and write the values ofa2, 2 xaxb, andb2in these columns. Step 2. Underline the unit digit ofb2(in Column III) and add its tens digit, if any, with 2 xaxb(in Column II). Step 3. Underline the unit digit in Column II and add the number formed by the tens and other ...
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Question: $\cot \theta \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}$ Solution: $\cot \theta \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}$ $=\cot \theta \cot \theta-\sec \left(90^{\circ}-\theta\r...
Read More →The displacement of a moving object in a given
Question: The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero ? Justify your answer. Solution: No. When object moves in a circular path of radius r, then displacement of object after completing a circle is zero but distance travelled $=2 \pi$....
Read More →In which of the following cases of motions,
Question: In which of the following cases of motions, the distance moved and the magnitude of displacement are equal? (a) If the car is moving on straight road (b) If the car is moving in circular path (c) The pendulum is moving to and fro (d) The earth is revolving around the Sun. Solution: (a) If the car is moving on straight road...
Read More →Slope of a velocity - time graph gives
Question: Slope of a velocity - time graph gives (a) the distance (b) the displacement (c) the acceleration (d) the speed. Solution: (c) Explanation : Slope of $v-t \mathrm{graph}=\frac{\text { change in velocity }}{\text { time }}=$ acceleration....
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