# Physics and Mathematics

## Concept Of Physics

### H C Verma

1   A vector $\vec{A}$ makes an angle of $20°$ and $\vec{B}$ makes an angle of $110°$ with the $X$-axis . The magnitudes of these vectors are $3 m$ and $4 m$ respectively. Find the resultant.

##### Solution :

The angle between $\vec{A}$ and $\vec{B}$ = $110^o$ - $20^o$ = $90^o$

therefore, Resultant vector makes angle $(53^o + 20^o)$ = $73^o$ with $x$-axis

Resultant R = $\sqrt{A^2 +B^2 + 2ABcos\theta}$ = $5m$

As shown in the figure ,

$\beta$ = $tan^{-1} \big(\frac{4sin90^o}{3+4 cos90^o})$ = $tan^{-1}$ $\big(\frac{4}{3})$ = $53^o$

$|\vec{A}|$ = $3m$ and $|\vec{B}|$ = $4m$

Let $\beta$ be the angle between $\vec{R}$ and $\vec{A}$

2   Let $\vec{A}$ and $\vec{B}$ be the two vectors of magnitude $10$ unit each. If they are inclined to the $X$-axis at angles $30°$ and $60°$ respectively, find the resultant.

##### Solution :

$|\vec{A}|$ and $|\vec{B}|$ = $10$ unit

$\beta$ be the angle between $\vec{R}$ and $\vec{A}$

Angle between $\vec{A}$ and $\vec{B}$ is $\theta$ = $60^o$ - $30^o$ = $30^o$

Therefore, Resultant makes $15^o$ + $30^o$ = $45^o$ angle with $x$-axis

R = $\sqrt{10^o + 10^o + 2.10.10.cos30^o}$ = $19.3$

$\beta$ = $tan^{-1} \big(\frac{10sin30^o}{10 + 10cos30^o})$ = $tan^{-1} \big (\frac{1}{2 + \sqrt{3}})$ = $tan^{-1} (0.26795)$ = $15^o$

3   Add vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ each having magnitude of $100$ unit and inclined to the $X$-axis at angles $45°$, $135°$ and $315°$ respectively.

##### Solution :

$x$ component of $\vec{A}$ = $100 cos 45^o$ = $\frac{100}{\sqrt{2}}$ unit

The resultant is $100$ unit at $45^o$ with $x$-axis

$y$ component of $\vec{B}$ = $100 cos 135^o$ = $\frac{100}{\sqrt{2}}$

Resultant = 100

$x$ component of $\vec{C}$ = $100 cos 315^o$ = $\frac{100}{\sqrt{2}}$

$\Rightarrow$ $\alpha$ = $tan^{-1}(1)$ = $45^o$

$y$ component of $\vec{A}$ = $100 cos 45^o$ = $\frac{100}{\sqrt{2}}$ unit

Resultant $Y$ component = $\frac{100}{\sqrt{2}}$ + $\frac{100}{\sqrt{2}}$ - $\frac{100}{\sqrt{2}}$ = $\frac{100}{\sqrt{2}}$

$x$ component of $\vec{B}$ = $100cos135^o$ = $\frac{100}{\sqrt{2}}$

$y$ component of $\vec{C}$ = $100 cos 315^o$ = $\frac{100}{\sqrt{2}}$

Tan $\alpha$ = $\frac{y\ component}{x\ component}$ = $1$

Resultant $x$ component = $\frac{100}{\sqrt{2}}$ - $\frac{100}{\sqrt{2}}$ + $\frac{100}{\sqrt{2}}$ = $\frac{100}{\sqrt{2}}$

4   Let $\vec{a}$ = $\vec{4 i}$ + $\vec{3j}$ and $\vec{b}$ = $\vec{3 i}$ + $\vec{4 j}$.

(a) $\vec{a}$ , (b) $\vec{b}$, (c) $\vec{a} + \vec{b}$ and (d) $\vec{a} - \vec{b}$ .

(a) Find the magnitudes of

##### Solution :

a) $|\vec{a}|$ = $\sqrt{4^2 + 3*2}$ = $5$

c) $|\vec{a}+ \vec{b}|$ = $|\vec{7i} + \vec{7j}|$ = $7\sqrt{2}$

$\vec{a}$ = $\vec{4i}$ + $\vec{3j}$ , $\vec{b}$ = $\vec{3i}$ + $\vec{4j}$

a) $|\vec{a} + \vec{b}|$ = $\sqrt{1^2 + (-1)^2}$ = $\sqrt{2}$

b) $|\vec{b}|$ = $\sqrt{9 + 16}$ = $5$

a) $\vec{a} - \vec{b}$ = $(-3 + 4)$ $\hat{i}$ + $(-4 + 3)\hat{j}$ = $\hat{i} - \hat{j}$

5   Refer to figure $(2-E1)$. Find (a) the magnitude, (b) $x$ and $y$ components and (c) the angle with the $X$-axis of the resultant of $\vec{OA}$, $\vec{BC}$ and $\vec{DE}$.

##### Solution :

$R_y$ = resultant $y$ component = $1$ + $1.3$ - $1$ = $1.3 m$

$x$ components of $\vec{BC}$ = $1.5 cos120^o$ = $-0.75$

$R_x$ = $x$ component of resultant = $\sqrt{3}$ - $0.75$ + $0$ = $0.98 m$

If it makes and angle $\alpha$ with positve $x$-axis

$y$ components of $\vec{OA}$ = $2sin30^o$ = $1$

$x$ components of $\vec{OA}$ = $2cos30^o$ = $\sqrt{3}$

$\Rightarrow$ $\alpha$ = $tan^{-1} 1.32$

$y$ components of $\vec{DE}$ = $1 sin270^o$ = $-1$

So, $R$ = Resultant = $1.6 m$

$x$ components of $\vec{DE}$ = $1cos270^o$ = $0$

Tan $\alpha$ = $\frac{y\ component}{x\ component}$ = $1.32$

$y$ components of $\vec{BC}$ = $1.5 sin120^o$ = $1.3$

6   Two vectors have magnitudes $3$ unit and $4$ unit respectively. What should be the angle between them if the magnitude of the resultant is (a) $1$ unit, (b) $5$ unit and (c) $7$ unit.

##### Solution :

Angle between them is $0^o$

a) If $R$ = $1$ Unit $\Rightarrow$ $\sqrt{3^2 + 4^2 + 2.3.4.cos\theta}$ = $1$

$\theta$ = $0^o$

b) $\sqrt{3^2 + 4^2 + 2.3.4.cos\theta}$ = $5$

$|\vec{a}|$ = $3m$ $|\vec{b}|$ = $4m$

c) $\sqrt{3^2 + 4^2 + 2.3.4.cos\theta}$ = $7$

$\theta$ = $180^o$

$\theta$ = $90^o$

7   A spy report about a suspected car reads as follows. "The car moved $2.00\ km$ towards east, made a perpendicular left turn, ran for $500\ m$, made a perpendicular right turn, ran for $4.00\ km$ and stopped". Find the displacement of the car.

##### Solution :

AD = $\sqrt{AE^2 + DE^2}$ = $6.02$ KM

$\theta$ = $tan^{-1}$ $\big(\frac{1}{12})$

$\vec{AD}$ = $2\hat{i}$ + $0.5\hat{j}$+ $4\hat{K}$ = $6\hat{i}$ + $0.5\hat{j}$

$Tan\theta$ = $\frac{DE}{AE}$ = $\frac{1}{12}$

The displacement of the car is $6.02$ km along the distance $tan^{-1}$ with positive $x$-axis

8   A carrom board ($4$ ft x $4$ ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole.

##### Solution :

$\Rightarrow$ 4 = 2$x$ = 4$x$

a) In $\Delta{ABC}$, AC = $\sqrt{AB^2 + BC^2}$ = $\frac{2}{3}\sqrt{10}$ ft

In $\Delta{ABC}$, $tan\theta$ = $\frac{x}{2}$ and in $\Delta{DCE}$, $tan\theta$ = $\frac{(2 - x)}{4}$ $tan\theta$ = $\frac{x}{2}$ = $\frac{(2 - x)}{4}$ = $4x$

$\Rightarrow$ 6$x$ = 4 $\Rightarrow$ $x$ = $\frac{2}{3}$ ft

9   A mosquito net over a $7$ ft x $4$ ft bed is $3$ ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the $X$-axis, its width as the Y-axis, and vertically up as the $Z$-axis, write the components of the displacement vector.

##### Solution :

a) magnitude of displacement = $\sqrt{74}$ ft

Here is the displacement vector $\vec{r}$ = $7\hat{i}$ + $4\hat{j}$ + $3\hat{k}$

b) the components of displacement vector are $7$ ft, $4$ ft and $3$ ft

10   Suppose $\vec{a}$ is a vector of magnitude 4.5 unit due north. What is the vector (a) $3\vec{a}$, (b)$- 4\vec{a}$?

##### Solution :

a) $3|\vec{a}|$ = $3$ x $4.5$ = $13.5$

$-4|\vec{a}|$ = $-4$ x $1.5$ = $-6$unit

$\vec{a}$ is a vector of magnitude $4.5$ unit due north.

$3\vec{a}$ is along north having magnitude $13.5$ units

$-4\vec{a}$ is a vector of magnitude $6$ unit due south.

11   Two vectors have magnitudes $2$ m and $3$ m. The angle between them is $60°$. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

##### Solution :

angle between them $\theta$ = $60^o$

$|\vec{a}$ x $\vec{b}|$ = $|\vec{a}|$ . $|\vec{b}|$$sin60^o = 2 x 3 x \sqrt\frac{3}{2} = 3\sqrt{3} m^2 |\vec{a}| = 2m , |\vec{b}| = 3m \vec{a} . \vec{b} = |\vec{a}| . |\vec{b}| cos60^o = 2 x 3 x \frac{1}{2} = 3$$m^2$

12   Let $A_{1}$ $A_{2}$ $A_{3}$ $A _{4}$ $A_{5}$ $A_{6}$ $A_{1}$ be a regular hexagon. Write the $x$-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that $cos0$ + $cos\frac{\pi}{3}$ + $cos\frac{2\pi}{3}$ + $cos\frac{3\pi}{3}$ + $cos\frac{4\pi}{3}$ + $cos\frac{5\pi}{3}$ = $0$. Use the known cosine values to verify the result.

##### Solution :

Here A = B = C = D = E = F (magnitude)

sin$\theta$ + sin$\frac{\pi}{3}$ + sin$\frac{2\pi}{3}$ + sin$\frac{3\pi}{3}$ + sin$\frac{4\pi}{3}$ + sin$\frac{5\pi}{3}$ = $0$

[As a resultant is zero. X component of resultant $R_x$ = $0$]

We know that according to polygon law addition, the resultant of these six vectors is zero.

Note: similarly it can be proved that,

So, Rx = A cos$\theta$ + A cos$\frac{\pi}{3}$ + A cos$\frac{2\pi}{3}$ + A cos$\frac{3\pi}{3}$ + A cos$\frac{4\pi}{4}$ + A cos$\frac{5\pi}{5}$ = $0$

= cos$\theta$ +cos$\frac{\pi}{3}$ +cos$\frac{2\pi}{3}$ + cos$\frac{3\pi}{3}$+ cos$\frac{4\pi}{3}$ + cos$\frac{5\pi}{3}$ = $0$

13   Let $\vec{a}$ = $2\vec{i}$ + $3\vec{j}$ + $4\vec{k}$ and $\vec{b}$ = $3\vec{i}$ + $4\vec{j}$ + $5\vec{k}$ . Find the angle between them.

a)

##### Solution :

14   Prove that $\vec{A}$ . ($\vec{A}$ x $\vec{B}$ ) = $O$.

##### Solution :

As $\vec{A}$ x $\vec{B}$ = $AB$ $sin\theta$ $\hat{n}$

Thus, $\vec{A}$ . ($\vec{A}$ X $\vec{B}$) = $0$

$\vec{A}$ . ($\vec{A}$ X $\vec{B}$) = $0$ (claim)

$AB$ $sin\theta$ $\hat{n}$ is a vector which is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ , that implies that it is also perpendicular to $\vec{A}$ . As dot product of two perpendicular vector is zero.

15   If $\vec{A}$ = $2\vec{t}$ +$3\vec{j}$ + $4\vec{k}$ and $\vec{B}$ = $4\vec{t}$ + $3\vec{j}$ + $2\vec{k}$ , Find $\vec{A}$ x $\vec{B}$.

##### Solution :

$\vec{A}$ x $\vec{B}$ = $\vmatrix{\hat{i}}&{\hat{j}}&{\hat{k}} \\ {2}&{3}&{4} \\ {4}&{3}7{2}}$

$\vec{A}$ = $2\hat{t}$ +$3\hat{j}$ + $4\hat{k}$ , $\vec{B}$ = $4\hat{t}$ + $3\hat{j}$ + $2\hat{k}$

16   If $\vec{A}$, $\vec{B}$, $\vec{C}$ are mutually perpendicular, show that $\vec{C}$ x ( $\vec{A}$ x $\vec{B}$ ) = $O$. Is the converse true ?

##### Solution :

$\vec{A}$ X $\vec{B}$ is vector which direction is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$

Therefore, angle between $\vec{C}$ and $\vec{A}$ x $\vec{B}$ is $0^o$ or $180^o$ (fig.1)

Given that $\vec{A}$, $\vec{B}$ and $\vec{C}$ are mutually perpendicular

Also $\vec{C}$ is perpendicular to $\vec{A}$ and $\vec{B}$

17   A particle moves on a given straight line with a constant speed $v$. At a certain time it is at a point P on its straight line path. $O$ is a fixed point. Show that $\vec{OP}$ x $\vec{v}$ is independent of the position $P$.

##### Solution :

From the figure,

It can be seen from the figure, $OQ$ = $OP$ $\sin\theta$ = $OP'$ $sin\theta$

the particle moves on the straight line $PP$' at speed $v$ .

Therefore, $\vec{OP}$ X $v$ is independent of the position $P$ .

$\vec{OP}$ x $v$ = $(OP)v$ $sin\theta\hat{n}$ = $v(OP)$ $sin\theta\hat{n}$ = $v(OQ)$ $\hat{n}$

So, whatever may be the position of the particle, the magnitude and direction of $\vec{OP}$ x $v$ remain constant.

18   The force on a charged particle due to electric and magnetic fields is given by $\vec{F}$ = $q$ $\vec{E}$ + $q$ $\vec{v}$ x $\vec{B}$. Suppose $\vec{E}$ is along the $X$-axis and $\vec{B}$ along the $Y$-axis. In what direction and with what minimum speed $v$ should a positively charged particle be sent so that the net force on it is zero ?

##### Solution :

$\Rightarrow$ $\vec{E}$ = -$(\vec{v}$ x $\vec{B})$ = 0

Again, $E$ = $vB$ $sin$ $\theta$ $\Rightarrow$ $v$ = $\frac{E}{B sin \theta}$

Give $\vec{F}$ = q$\vec{E}$ + q$(\vec{v}$ x $\vec{B})$ = 0

So, the practicle must be projected at a minimum speed of $\frac{E}{B}$ along +ve $z$-axis ($\theta$ = $90^o$) as shown in the figure, so that the force is zero.

So, direction of $\vec{v}$ x $\vec{B}$ should be opposite to the direction of $\vec{E}$ . Hence, $\vec{v}$ should be in the positive $yz$-plane.

For $v$ to be minimum, $\theta$ = $90^o$ and so $v_{min}$ = $\frac{F}{B}$

19   Give an example for which $\vec{A}$ . $\vec{T}$ = $\vec{C}$ but $\vec{A}$ $\ne$ $\vec{C}$.

##### Solution :

$\vec{B}$ $\cdot$ $\vec{C}$ = 0

$\vec{A}$ $\bot$ $\vec{B}$ . $\vec{B}$ along west

EQAE

$\vec{C}$ along north

For example, as shown in the figure,

Therefore, $\vec{A}$ $\cdot$ $\vec{B}$ = $\vec{B}$ $\cdot$ $\vec{C}$

But $\vec{B}$ $\ne$ $\vec{C}$

$\vec{B}$ $\bot$ $\vec{C}$ . $\vec{A}$ along south

$\vec{A}$ $\cdot$ $\vec{B}$ = 0

20   Draw a graph from the following data. Draw tangents at $x$= $2$, $4$, $6$ and $8$. Find the slopes of these tangents. Verify that the curve drawn is $y$ = $2x^2$ and the slope of tangent is $tan$$\theta = \frac{dy}{dx} = 4x. y 2 8 18 32 50 72 98 128 162 200 x 1 2 3 4 5 6 7 8 9 10 ##### Solution : To find slope at any point, draw a tangent at the point and extend the line to meet x-axis . Then find tan \theta as shown in the figure. Slope = tan \theta = \frac{dy}{dx} = \frac{d}{dx} (2x^2) = 4x The graph y = 2x^2 should be drawn by the student on a graph for exact results. It can be checked that, 21 A curve is represented by y = sin x. If x is changed from \frac{\pi}{3} to \frac{\pi}{3} + \frac{\pi}{100} . Find approximately the change in y. ##### Solution : So, y + \Delta{y} = Sin (x + \Delta{x}) = \big(\frac{\pi}{3} + \frac{\pi}{100}) - sin \frac{\pi}{3} = 0.0157 . y = sin x \Delta{y} = sin (x + \Delta{x}) - sin x 22 The electric current in a charging R-C circuit is given by i = i_0 , e^\frac{-t}{RC} where i_o , R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = RC , (c) t = 10 RC. ##### Solution : when, b) When t = RC , \frac{di}{dt} = \frac{-i}{RCe} Given that, i = i_0e^\frac{-t}{RC} \\ \therefore Rate of change of current = \frac{di}{dt} = \frac{d}{dt} i_0e^\frac{-i}{RC} = I_0\frac{d}{dt} e^\frac{-t}{RC} = \frac{-i_0}{RC} x e^\frac{-t}{RC} a) t= 0 , \frac{di}{dt} = \frac{-i}{RC} c) When t = 10 RC , \frac{di}{dt} = \frac{-i_0}{RCe^10} 23 The electric current in a discharging R - C circuit is given by i = i_o , e^\frac{-t}{RC} where i_o , R and C are constant parameters and t is time. Let i_o = 2\cdot00 A, R = 6\cdot00 x 10^5 \Omega and C = 0\cdot500 \mu$$F$ (a) Find the current at $t$ = $0\cdot3$ s. (b) Find the rate of change of current at $t$ = $0\cdot3$ s. (c) Find approximately the current at $t$ = $0\cdot31$ s.

$\frac{$6.67$x$10^{-11}$x$10^{6}$}{6.64 x 10^{-17}}$

##### Solution :

Equation $i$ = $i_0e^\frac{-t}{RC}$ $\\$ $I_0$ = $2A$, R = $6$ x $10^{-5}$ $\Omega$ , C = $0.0500$ x $10{-8}$ $F$ = $5$ x $10^{-7}$ $F$ $\\$ a) $i$ = $2$ X $e\frac{-0.3}{6\times 0^3\times 5\times 10^{-7}}$ = $2$ x $e\frac{-0.3}{0.3}$ = $\frac{2}{e}$ $amp$ . $\\$ b) $\frac{di}{dt}$ = $\frac{-i_0}{RC}$ $e\frac{-t}{RC}$ when, $t$ = $0.3$ sec $\Rightarrow$ $\frac{di}{dt}$ = - $\frac{2}{0.30}$ $e(\frac{-0.3}{0.3})$ = $\frac{-20}{3e}$ Amp/sec $\\$ c) At $t$ = $0.31$ sec, $i$ = $2e^(\frac{-0.3}{0.3})$ = $\frac{5.8}{3e}$ Amp

24   Find the area bounded under the curve $y$ = $3x6^2$ + $6x$ + $7$ and the $X$-axis with the ordinates at $x$ = $5$ and $x$ = $10$.

##### Solution :

$\therefore$ Area bounded by the curve, $x$ axis with coordinates with $x$ = 5 and $x$ = 10 is given by,

y = 3$x^2$ + 6$x$ + 7

Area = $\int^{y}_{0}$ $dy$ = $\int^{10}_{5}$ (3$x^2$ + 6$x$ + 7)$dx$ = 3$\frac{x^3}{3}$ $\big]^{10}_{5}$ + 5$\frac{x^2}{3}$ $\big]^{10}_{5}$ + 7$x$ $\big]^{10}_{5}$ = 1135 sq.units

25   Find the area enclosed by the curve $y$ = $sin$ $x$ and the $X$-axis between $x$ = $0$ and $x$ = $n$ .

##### Solution :

Area = $\int^{y}_{0}$ $dy$ = $\int^{x}_{0}$ sin$xdx$ = - $[cos x ]^{\pi}_{0}$ = 2

26   Find the area bounded by the curve $y$ = $e^{-x}$, the $X$-axis and the $Y$-axis.

##### Solution :

When $x$ = 0 , $y$ = $e^{-0}$ = 1

So, the required area can be found out by integrating the function from 0 to $\infty$ .

The given function is $y$ = $e^{-x}$

$X$ increases , $Y$ value decreases and only at $x$ = $\infty$ , $y$ = 0

So , Area = $\int_{0}^{\infty}$ $e^{-x}dx$ = -$[e^{-x}]^{\infty}_{0}$ = 1

27   A rod of length $L$ is placed along the $X$-axis between $x$ = $0$ and $x$ = $L$. The linear density (mass/length) $\rho$ of the rod varies with the distance $x$ from the origin as $\rho$ = $a$ + $bx$. (a) Find the S.I units of $a$ and $b$. (b) Find the mass of the rod in terms of $a$, $b$ and $L$

##### Solution :

a0 S.I. unit of '$a$' = $\frac{kg}{m}$ and S.I. unit of '$b$' = $\frac{kg}{m^2}$

$\therefore$ $dm$ = mass of the element = $\rho$ $dx$ = ( a +bx ) dx

$\rho$ = $\frac{mass}{length}$ = a + bx

b) Let us consider a small element of length '$dx$' at a distance $x$ fro, the orgin as shown in the figure.

So, mass of the rod = $m$= $\int$ $dm$ = $|int_{0}^{L}$ ( a + bx ) dx = $\big[ ax + \frac{bx^2}{2}\big]_{0}^{L}$

28   The momentum $p$ of a particle changes with time $t$ according to the relation $\frac{dp}{dt}$ = ($10 N$) + ($2\frac{N}{s}$)$t$ . If the momentum is zero at $t$ = $0$, what will the momentum be at $t$ = $10$ $s$

##### Solution :

momentum is zero at $t$ = 0

$dp$ = $[ (10 N) + 2Nst] dt$

$\frac{dp}{dt}$ = (10 N) + (2 $\frac{N}{S}$) t

$\therefore$ momentum at $t$ = 10 sec will be

$\int_{0}^{p}$ $dp$ = $\int_{0}^{10}$ $(10dt)$ + $\int_{0}^{10}$ $(2tdt)$ = $10t]_{0}^{10}$ + 2 $\frac{t^2}{2}\big]_{0}^{10}$ = 200 kg $\frac{m}{s}$

29   The changes in a function $y$ and the independent variable $x$ are related as $\frac{dy}{dx}$ = $x^2$ . Find $y$ as a function of $x$ .

##### Solution :

$\Rightarrow$ $dy$ = $x^2$ $dx$

$\int$ $dy$ = $\int$ $x^2$ $dx$ $\Rightarrow$ $y$ = $\frac{x^3}{3}$ + c

The change in a function of $y$ and the independent variable $x$ are related as $\frac{dy}{dx}$ = $x^2$ .

Taking integration of both sides,

$\therefore$ $y$ as a function of $x$ represented by $y$ = $\frac{x^3}{3}$ + c

30   Write the number of significant digits in (a) $1001$, (b) $100\cdot1$, (c) $100\cdot10$, (d) $0\cdot001001$.

##### Solution :

d) 0.001001 No. of significant digits = 4

a) 1001

No. of significant digits = 5

b) 100.1

The number significant digits

c) 100.10

No. of significant digits = 4

No. of significant digits = 4

31   A metre scale is graduated at every millimetre. How many significant digits will be there in a length measurement with this scale ?

##### Solution :

1 m = 100 mm

So, the no. of significant digits may be 1, 2, 3, or 4.

The metre scale is graduated at every millimeter.

The maximum no. of significant digit may be 1 ( e.g. for measurements like 5 mm, 7 mm etc) and the maximum no. of significant digits may be 4 (e.g. 1000 mm)

32   Round the following numbers to $2$ significant digits. (a) $3472$, (b) $84\cdot16$, (c) $2\cdot55$ and (d) $28\cdot5$.

##### Solution :

So, the next two digits are neglected and the value of 4 is increased by 1

b) value = 84 $\\$ c) 2.6 $\\$ d) value is 28.

In the value 3472, after the digit 4, 7 is present. Its value is greater than 5.

$\therefore$ value becomes 3500

33   The length and the radius of a cylinder measured with a slide callipers are found to be $4\cdot54$ cm and $1\cdot75$ cm respectively. Calculate the volume of the cylinder.

##### Solution :

Length = $I$ = 4.54 cm, radius = $r$ = 1.75 cm

Since, the minimum mo. of significant digits on a particular term is 3, the result should have 3 significant digits and others rounded off.

Given that, for the cylinder

Since, it is to be rounded off to 3 significant digits, $V$ = $43.7$ $cm^3$

Volume = $\pi r^2 I$ = $\pi$ x $(4.54)$ x $(1.75)^2$

So, volume $V$ = $\pi r^2 I$ = $(3.14)$ x $(1.75)$ x $(1.75)$ x $(4,54)$ = $43.6577$ $cm^3$

34   The thickness of a glass plate is measured to be $2\cdot17$ mm, $2\cdot17$ mm and $2\cdot18$ mm at three different places. Find the average thickness of the plate from this data.

##### Solution :

35   The length of the string of a simple pendulum is measured with a metre scale to be $90\cdot0$ cm. The radius of the bob plus the length of the hook is calculated to be $2\cdot13$ cm using measurements with a slide callipers. What is the effective length of the pendulum ? (The effective length is defined as the distance between the point of suspension and the centre of the bob.)

##### Solution :

Actual effective length = (90.0 + 2.23) cm

So, the addition must be done by considering only 2 significant digits of each measurement.

As shown in the figure,

But, in the measurement 90.0 cm, the no. of significant digits is only 2.

So, effective length = 90.0 + 2.1 = 92.1 cm

36   The thickness of a glass plate is measured to be 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data.

##### Solution :

Average thickness = $\frac{2.17 + 2.17 + 2.18}{3}$ = 2.1733 mm

We know that,

Rounding off to 3 significant digits, average thickness = 2.17 mm

37   The length of the string of a simple pendulum is measured with a metre scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated to be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum ? (The effective length is defined as the distance between the point of suspension and the centre of the bob.)

##### Solution :

38   DHTDFJHG $\\$ A $\Omega$ $\\$ $\omega$ $\\$ $\vec{A} = 2^{2}i + R_{th}$ $\\$ $y = x^{2} + 2x+ 1$ $\quad$ ... (1) $\\$ The left side of the equation describes acceleration, and may be composed of time-dependent and convective components (also the effects of non-inertial coordinates if present). $\vec{A} = 2^{2}i + R_{th}$ $\\$ The right side of the equation is in effect a summation of hydrostatic effects, the divergence of deviatoric stress and body forces (such as gravity).

JGFJH

##### Solution :

zczcdsccs

39   1. A particle executes simple harmonic motion with an amplitude of $10$ cm and time period $6$ s. At $t =0$ it is at position $x= 5$ cm going towards positive x-direction. Write the equation for the displacement $x$ at time $t$. Find the magnitude of the acceleration of the particle at $t = 4$ s

##### Solution :

therefore Equation of displacement $x$ = ($10cm$) sin $\big(\frac{\pi}{3}\big)$

At $t =0$, $x = 5cm$

sin $\phi$ = $1/2$ $\Rightarrow$ $\phi$ = $\frac{\pi}{6}$

$x$ = $10$ sin $\big[\frac{\pi}{3}$ x $4$ + $\frac{\pi}{6}\big]$ = $10$ sin $\big[\frac{8 \pi + \pi}{6}\big]$

so, $w$ = $\frac{2\pi}{T}$ = $\frac{2\pi}{6}$ = $\frac{\pi}{3}$ $sec^{-1}$

Given, $r = 10cm$.

Acceleration a = -$w^2$x = -$\big(\frac{\pi^2}{9}\big)$x(-10) = 10.9 $\approx$ 0.11 cm/sec

so, $5 = 10$ sin($w$ x $0$ x $\phi$) = $10$ sin $\phi$ $\\$ [$y$ = $r$ $sin$ $wt$]

(ii) At $t$ = $4$ second

T = 6 sec

= $10$ sin $\big(\frac{3\pi}{2}\big)$ = $10$ sin $\big(\pi+\frac{\pi}{2}\big)$ = -$10$ sin $\big(\frac{\pi}{2}\big)$ = -$10$

At, $t = 0$,$x = 5 cm$