Grade 11 NEB Practice set I with solutions | Physics | Physics in Depth

Physics Practice Set I

1. Can a quantity have unit but no dimensions? If yes give an example.
Solution :
Yes, a quantity can have unit but still be dimensionless. For example, the unit of angle is radian which has no dimension.


2. Can the resultant of three forces be zero? Explain.
Solution :
Force is a vector quantity and has specific rules of addition and subtraction. The resultant of three forces can be zero if both of the following conditions are satisfied:

• The magnitude of resultant of two forces is exactly equal to the magnitude of the third force.
• The direction of the resultant of these two forces is exactly opposite to the direction of third force.



3. Is there any point for projectile at which its acceleration is perpendicular to its velocity?
Solution :
Consider a projectile fired at an angle with the horizontal or the levelled ground. When the projectile reaches the maximum height, it has no vertical velocity and in this case, the acceleration due to gravity is perpendicular to the horizontal component of the velocity of projectile. Thus, maximum height is the point for projectile at which its acceleration is perpendicular to its velocity.


4. Explain the significance of the banking of a curved road?
Solution :
When the vehicle moves in the level curved path, the necessary centripetal force is provided by the friction between the tyres and the road. Thus, to increase the centripetal force, friction should be increased which causes the wear and tears of the tyres. Also, the maximum speed with which the vehicle can take a circular turn is $v=\sqrt{\mu rg}$ where $\mu$ is the coefficient of friction, r is the radius of circular path and g, the acceleration due to gravit. If the velocity of the vehicle is greater than this, the skidding of the vehicle occurs and it moves out of the road. However, when the roads are banked, the horizontal component of the normal reaction on the tyres provides the necessary centripetal force. This illustrates the significance of the banking of a curved road.


5. Why does glass of windows are possible to be cracked during cold winter?
Solution :
During winter, the temperature difference between the inside and outside of the room is unequal becaue glass is a bad conductor of heat. Due to this, there is unequal expansion between inner and outer layers of the windows glass. This develops the thermal stress. Larger the temperature difference, larger is the thermal stress developed. Hence, due to this reason, the glass of windows are possible to be cracked during cold winter.


6. We feel metal door colder than the wooden door even when both of them are at same temperature. Why?
Solution :
Metal door is good conductor of heat and wooden door is a bad conductor of heat. When we touch the metal door, heat flows from our body to the metal door and we feel cold. On the other hand, no heat flows from our body to the wooden door and hence it does not appear cold though being at same temperature as of the metal door.


7. Why there is no dispersion of monochromatic light?
Solution :
For the dispersion of light, light should have different wavelengths. White light has this characteristic i.e., it is the combination of seven visible colors with different wavelengths. But, monochromatic light has a particular single wavelength. Due to this reason, no dispersion occurs of monochromatic light.


8. A convex lens is immersed in water; will its focal length change? Explain.
Solution :
From the lens maker's formula;
$$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}+\frac{1}{R_2})$$ In air, $$\frac{1}{f_a}={(}^a{\mu }_g-1)(\frac{1}{R_1}+\frac{1}{R_2})...(i)$$ where ${}_a{\mu }_g$ is refractive index of glass with respect to air , $f_a$ is the focal length of lens in air, $R_1$ and $R_2$ are the radii of curvature of two surface of the lens.
When the convex lens is immersed in water,
$$\frac{1}{f_w}={(}^w{\mu }_g-1)(\frac{1}{R_1}+\frac{1}{R_2})...(ii)$$ where ${}_w{\mu }_g$ is refractive index of glass with respect to air , $f_w$ is the focal length of lens in air, $R_1$ and $R_2$ are the radii of curvature of two surface of the lens.
Dividing equation (i) by (ii),
$$\begin{array}{rl}\frac{f_w}{f_a}& =\left(\frac{{}^a{\mu }_g-1}{{}^w{\mu }_g-1}\right)\\ & =\frac{{}^a{\mu }_g-1}{(\frac{{}^a{\mu }_g}{{}^a{\mu }_w}-1)}\\ & =\frac{1.5-1}{\frac{1.5}{1.33}-1}\\ \therefore \frac{f_w}{f_a}& =3.9\end{array}$$ Thus, $f_w=3.9f_a$ Hence, the focal length of convex lens increases by 3.9 times when dipped in water.


9. What do you mean by electron volt (eV)? Explain.
Solution :
The potential difference is expressed as,
$$V=\frac{W}{q}$$ where, W is the work done by the charge q .
When V = 1 V and q = 1.6 $\times$ 10${}^{-19}$ C, $$W=1.6\times {10}^{-19}\text{J}$$ It is the expression for the energy gained by the electron when it is accelerated through 1 volt of potential difference. This energy is known as electron volt (eV). Thus, the energy gained by an electron which has been accelerated through a potential difference of one volt is called an electron volt (eV).
$$\therefore 1\text{eV}=1.6\times {10}^{-19}\text{J}$$
10. Differentiate between dielectric constant and dielectric strength.
Solution :
Dielectric constant (K) is defined as the ratio of the permittivity of the substance to that of free space i.e., $K=\frac{ϵ}{{ϵ}_0}$. It can also be defined as the ratio of the capacitance of the capacitor having a dielectric between the plates to the capacitance of the same capacitor without any dielectric i.e., $K=\frac{C}{C_0}$.
Dielectric strength of the dielectric, on the other hand, is the value of electric field intensity that can be applied to the dielectric without its electric breakdown. Electric breakdown is the condition when the dielectric becomes conducting.


11. What types of mirror will you suggest for saving or makeup purpose and why?
Solution :
A convex mirror produces an erect diminished image. But, the concave mirror produces an erect and enlarged image when the distance from the object to the mirror is less than the focal length of the mirror. Thus, I will suggest concave mirror for saving or make up purpose as it produces enlarged image.


12. Why does the sky appear blue? Explain.
Solution :
According to the Rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of the wavelength. This means, more the wavelength , less is the scattering and less the wavelength , more is the scattering. In a visible spectrum, the white light component having lesser wavelength is blue because of which it scatters more. Thus, sky appears blue.


13. Can two electric lines of force intersect each other? Explain.
Solution :
The tangent drawn on the electric lines of force gives the direction of force at that point due to the point charges. If two electric lines of force intersect each other, there will be two tangents showing two direction of force at the same point which is not possible. Thus, two electric lines of force cannot intersect each other.


14. A cricket ball of mass 145 gm is moving with a velocity of 14 m/s and is being hit by a bat, so that the ball is turned back with a velocity of 22 m/s. The force of blow acts on the ball for 0.015 sec. Find the average force exerted by the bat on the ball.
Solution :
mass of cricket ball, $m$ = 145 gm = 0.145 kg
initial velocity, $v_1$ = 14 m/s
final velocity, $v_2$ = -22 m/s ; negative sign is to indicate the opposite direction of velocity to the initial.
time of impact, $t$ = 0.015 sec
average force exerted by the bat on the ball, $F_{avg}$ = ?
We know, $$\begin{array}{rl}& =\frac{m(v_2-v_1)}{t}\\ & =\frac{0.145\times (-36)}{0.015}\\ & =-348\text{N}\end{array}$$ Magnitude of average force = 348 N.
Conclusion : Thus, the required solution is 348 N.


15. An object of mass 10kg is whirled round a horizontal circle of radius 4m by a revolving string inclined to the vertical. If the uniform speed of the object is 5m/s, calculate (i) the tension in the string, (ii) the angle of inclination of the string to the vertical.
Solution :
mass of object, $m$ = 10 kg
radius of horizontal circle, $r$ = 4 m
speed of object, $v$ = 5 m/s
tension in the string, $T$ = ?
angle of inclination of the string to the vertical, $\theta$ = ?
For horizontal circle,
$$\begin{array}{rl}Tcos\theta & =mg...(i)\\ Tsin\theta & =\frac{m{v}^2}{r}...(ii)\end{array}$$
Dividing equation (ii) by (i),
$$\begin{array}{rl}\frac{Tsin\theta }{Tcos\theta }& =\frac{\frac{m{v}^2}{r}}{mg}\\ tan\theta & =\frac{v^2}{rg}\\ tan\theta & =\frac{5^2}{4\times 9.8}\\ \theta & =ta{n}^{-1}0.64\\ \theta & ={32}^{\circ }\end{array}$$
Now substituting this value in equation (i) to get T, $$\begin{array}{rl}Tcos\theta & =mg\\ T\times cos32& =10\times 9.8\\ T& =\frac{98}{0.84}\\ \therefore T& =116.67\text{N}\end{array}$$
Conclusion : Thus, the required solutions are (i) 116.67 N and (ii) 32 ${}^{\circ }$.


16. A vehicle of mass 2000 kg travelling at 10 m/s on a horizontal surface is brought to rest in a distance of 12.5 m by the action of its breaks. Calculate the average retarding force. What power must the engine develop in order to take the vehicle up an incline of 1 in 10 at a constant speed of 10m/s? If the frictional resistance is equal to 200 N?
Solution :
mass of vehicle, $m$ = 2000 kg
For the first part,
Initial speed, $u$ = 10 m/s
Final speed, $v$ = 0 $\because$ the vehicle is stopped.
distance travelled, $s$ = 12.5 m
average retarding force, $F_r$ = ?

Now, using the equation of motion, $$\begin{array}{rl}{v}^2& =u^2+2as\\ 0& ={10}^2+2\times a\times 12.5\\ a& =-4m/s^2\end{array}$$ Thus the retardation is 4 m/s${}^2$ and hence the average retarding force is, $$F_r=2000\times 4=8000\text{N}$$

For the second part,
sin$\theta$= $\frac{1}{100}$
speed, $v$ = 10 m/s
frictional force, $f$ = 200 N
Let $F$ be the force of engine of the car.

When the vehicle is moved up the incline with the constant speed, the net force is zero.
Thus,
$$\begin{array}{rl}\text{net force}& =F-(f+mgsin\theta )\\ 0& =F-(200+2000\times 10\times \frac{1}{100})\\ F& =2200\text{N}\end{array}$$

Thus the power developed by the engine is, $$P=F\times v=2200\times 10=22000\text{W}$$
Conclusion : Thus the required solutions are 8000 N and 22000 W respectively.


17. A uniform steel wire 3m long weighing 21 g is extended by a load of 2.5 kg. Calculate (i) the extension produced and (ii) the energy stored in the wire. [Density of steel = 7.8 x 10${}^3$ kg/m${}^3$; Y= 2x10${}^{11}$ N/m${}^2$; g = 10 m/s${}^2$.]
Solution :
length of steel wire, $L$ = 3 m
mass of wire, $m$ = 21 g = 0.021 kg
density of steel, $\rho$ = 7.8 x 10${}^3$ kg/m${}^3$
Young's modulus of elasticity, $Y$ = 2x10${}^{1}1$ N/m${}^2$
$g$ = 10 m/s${}^2$
(i) extension produced , $l$ = ?
(ii) energy stored, $E$ = ?
load = 2 kg
force applied on wire, $F$ = 2 $\times$ 10 = 20 N

We have, $$F=\frac{YAl}{L}$$ Here, $$A=\frac{m}{L\times \rho }=\frac{0.021}{3\times 7800}=8.9\times {10}^{-7}{m}^2$$ $\because$ V = A $\times$ L and m = $\rho$ $\times$ V Thus, $$\begin{array}{rl}l& =\frac{FL}{YA}\\ & =\frac{60}{2\times {10}^{11}\times 8.9\times {10}^{-7}}\\ \therefore l& =3.4\times {10}^{-4}\text{m}\end{array}$$ Now the energy stored is given by the relation, $$\begin{array}{rl}E& =\frac{1}{2}\times F\times l\\ & =\frac{1}{2}\times 20\times 3.4\times {10}^{-4}\\ \therefore E& =3.4\times {10}^{-3}\text{J}\end{array}$$

Conclusion : Thus the required solutions are 3.4 $\times$ 10${}^{-4}$ m and 3.4 $\times$ 10${}^{-3}$ J respectively.


18. What is the result of mixing 20 gm of water at 90${}^{\circ }$ C with 10 gm of ice at -10 ${}^{\circ }$ C? [Specific heat capacity of ice = 0.5 cal gm${}^{-1}$ ${}^{\circ }$C${}^{-1}$, Specific heat capacity of water = 1 cal gm${}^{-1}$ ${}^{\circ }$ C${}^{-1}$ ,latent heat of fusion of ice = 80 cal gm${}^{-1}$].
Solution :
( In this type of questions, you have to check whether all ice melts or not at first before proceeding. )
mass of water, $m_w$ = 20 g
mass of ice, $m_i$ = 10 g
temperature of water, ${\theta }_w$ = 90${}^0$ C
temperature of ice, ${\theta }_i$ = -10${}^0$ C
specific heat of ice, $s_i$ = 0.5 cal/g
specific heat of water , $s_w$ = 1 cal/g
Check!
Amount of heat required to convert 10 g of ice at -10${}^0$ C to ice at 0${}^0$ C is,
$$\begin{array}{rl}& =m_i{s}_i[0-(-10)]\\ & =10\times 0.5\times 10\\ & =50\text{cal}\end{array}$$ Amount of heat required to convert ice at 0${}^0$ C ice water at 0${}^0$ C is, $$\begin{array}{rl}& =m_i{L}_i\\ & =10\times 80\\ & =800\text{cal}\end{array}$$ Amount of heat given out when 20 g water cools from 90 ${}^{\circ }$ C to 0${}^{\circ }$ C is, $$\begin{array}{rl}& =m_w{s}_w(90-0)\\ & =20\times 1\times 80\\ & =1800\text{cal}\end{array}$$ As the heat given out by water when it cools from 80 ${}^{\circ }$ C to 0${}^{\circ }$ C i.e., 1800 $\text{cal}$ is greater than that required to convert ice at -10 ${}^{\circ }$ C to water at 0${}^{\circ }$ C i.e., 85000 cals, so all ice melts.
Proceed!
Let $\theta$ be the final temperature of the mixture.
Amount of heat gained by ice from -10${}^0$ C to ice at 0${}^0$ C = 50 cal.
Amount of heat gained by ice from 0${}^0$ C to water at 0${}^0$ C = 800 cal.
Amount of heat gained by water (ice) at 0${}^0$ C to $\theta$ ${}^{\circ }$ C = $m_i$ $\times$ L_i $\times$ $(\theta -0)$ = 10 $\theta$
Amount of heat lost by water at 90 ${}^{\circ }$ C to water at $\theta$ ${}^{\circ }$ C = $m_w$ $\times$ $s_w$ $\times$ (90-$\theta$) = 1800 - 20 $\theta$

From the principle of calorimetry, $$\begin{array}{rl}\text{heat loss}& =\text{heat gain}\\ 1800-20\theta & =850+10\theta \\ 1800-850& =30\theta \\ \theta & =\frac{950}{30}\\ \therefore \theta & ={31.67}^{\circ }\text{C}\end{array}$$

Conclusion : Resulting temperature = 31.67 ${}^{\circ }$ C and all the ice melts such that the mass of water becomes 30 g.


19. The root mean square (rms) speed of a gas molecule is 600 ms${}^{-1}$ at 50 ${}^{\circ }$ C. Calculate the rms speed of the gas at 100 ${}^{\circ }$ C.
Solution :
temperature, $T_1$ = 273 + 50 = 323 K
rms speed, $(C_{rms}{)}_1$ = 600 m/s
temperature, $T_2$ = 273 + 100 = 373 K
rms speed, $(C_{rms}{)}_2$ = ?

We know that, $$C_{rms}\propto \sqrt{T}$$ So, $$\begin{array}{rl}\frac{(C_{rms}{)}_1}{(C_{rms}{)}_2}& =\sqrt{\frac{T_1}{T_2}}\\ \frac{600}{(C_{rms}{)}_2}& =\sqrt{\frac{323}{373}}\\ (C_{rms}{)}_2& =600\times \sqrt{\frac{373}{323}}\\ \therefore (C_{rms}{)}_2& =644.77\text{m/s}\end{array}$$
Conclusion : Thus, the required solution is 644.77 m/s.


20. The element of an electric heater with an output 1.5 KW is a cylinder of 30 cm long and 0.2 cm in diameter. Calculate its temperature if it behaves as a black body. [Stefan’s constant = 5.7 x 10${}^{-8}$ Wm${}^{-2}$K${}^{-4}$]
Solution : Click here! and scroll down to q.no. 6 of numerical problem to find the solution.


21. In order to correct his near point to 25 cm a man is given spectacles with converging lenses of 50 cm focal length and to correct his far point to infinity, he is given diverging lenses of 200 cm focal length. Determine his far and near points when not wearing the spectacles.
Solution :
The lens formula is given by, $$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$$ For correcting near point to be 25 cm,
u = 25 cm
f = 50 cm
Using lens formula, $$\begin{array}{rl}\frac{1}{50}& =\frac{1}{25}+\frac{1}{v}\\ \frac{1}{50}-\frac{1}{25}& =\frac{1}{v}\\ \therefore v& =-50\text{cm}\end{array}$$ This gives the near point of a man without wearing spectacles.
For far point,
u = $\mathrm{\infty }$
f = -200 cm
Using lens formula as above, you will obtain v = -200 cm.
This gives the far point of a man without wearing spectacles.
Conclusion : The required solutions are -50 cm and -200 cm.


22. Two point charges +1 μC and +4 μC are placed at a distance of 0.12m apart. Determine the point of the line joining two charges where net force acting on the unit positive charge is zero.
Solution :
$q_1$ = +1 $\mu$ C = 10${}^{-6}$ C
$q_2$ = +4 $\mu$ C = 4 $\times$ 10${}^{-6}$ C

Let the null point be at distance x from bigger charge and (0.12-x) from smaller charge.
At the null point, Electric field intensity due to $q_1$ i.e., $E_1$ is equal to the electric field intensity due to $q_2$ i.e., $E_2$.

Thus, $$\begin{array}{rl}{E}_1& =E_2\\ \frac{q_1}{4\pi {ϵ}_0{x}^2}& =\frac{q_2}{4\pi {ϵ}_0(0.12-x{)}^2}\\ \frac{1}{x^2}& =\frac{4}{(0.12-x{)}^2}\end{array}$$ I have left this calculation as your workout! Hint : x = 0.04 m.


23. A wire strain gauge with a gauge factor K = 2 is bonded to a steel member which is subjected to a strain of 10^-6. If original no-strain resistance of the gauge is 120 Ω, calculate the change in gauge resistance. Solution :
    gauge factor, K = 2
    strain, σ = 10^-6
    initial resistance, R = 120 Ω
    change in gauge resistance, ΔR =?
    
    Now,
    $$\begin{array}{rl}K& =\frac{\frac{\mathrm{\Delta }R}{R}}{\sigma }\\ \mathrm{\Delta }R& =\sigma KR\\ & ={10}^{-6}\times 2\times 120\\ \therefore \mathrm{\Delta }R& =240\times {10}^{-6}\mathrm{\Omega }\end{array}$$ Conclusion : The change in gauge resistance is 240 × 10^-6 Ω .