67486 + Interview Questions in Secondary School in Physics

1.	How does jet aeroplane fly in air
Answer» A plane's engines aredesigned to move it forward at high SPEED. ... The force of the HOT EXHAUST gas shooting backward from the jet engine pushes the plane forward. That CREATES a MOVING current of air over the wings.

Discussion

2.	Two fat astronauts each of mass 120 kg are traveling in a closed spaceship moving at a speed of 15 km/s in the outer space far removed from all other material objects. The total mass of the spaceship and its contents including the astronauts is 660 kg. If the astronauts do slimming exercise and thereby reduce their masses to 90 kg each with ,what velocity will the spaceship move?
Answer» $\fbox{Heya!!}$ Given Data :- Speed of space ship :- 15km/s Mass of astronauts :- 120 +120 = 240 Mass of space ship :- 660 including mass of astronauts IF mass of astronauts is REDUCED to 90 kg 240 - 90 =150 so 150 kg mass would be reduced total mass would alter as 660-150 = 510 NOW THE QUESTION IS WHAT WOULD BE THE VELOCITY OF SPACE SHIP $answer$ → velocity would be constant , There would be No Change in velocity WHY ?? • First we should now what will CAUSE a change in velocity ◾ If any force ACTS on the body in OPPOSITE direction then the velocity change HERE IN THE GIVEN QUESTION » Mass had changed but there is no force Created there » Since there is no force created , there would be no change in velocity Note :- At first I have did calculations to show the change in mass , and therefore to conclude that No Force would be created ========================== Glad if Helped !! Regards :- SunitaWilliams

Discussion

3.	Why a cricketer lower his hand while catching a ball?
Answer» The FIELDER fielder lowers his hand while catching the ball to increase the TIME period so that the rate of chang of momentum will be LESS the FORCE will also be APPLIED less on the fielder's hand.

Discussion

4.	Plz answer the 14th question if possible...I guess it's incomplete
Answer» BRO. PHOTO is not VISIBLE prperly

Discussion

5.	the velocity of a particle any time t is given by v=\|t-2\| m/s. The average velocity between t=0 to t=3 second is
Answer» Hey!!!...Here is ur ANSWER At TIME t, v=\|t-2\| m/s at t=0 , v=\|0-2\| => v=2 m/s at t=3 , v=\|3-2\| => v= 1m/s Average VELOCITY = (2+1)/2 => 3/2 m/s Hope it will help you

Discussion

6.	A force of 25 N act on a mass 500g resting on frictionless surface what is the acceleration produce?
Answer» M=500×1/1000=0.5 f=ma 25= 0.5 ×a 25/0.5=a 50m/s^2 =a

Discussion

7.	Question 5.15: A short bar magnet of magnetic moment 5.25 × 10 −2 J T −1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-202
Answer» (a) at NORMAL bisector, $\bf{tan\theta=\frac{B_H}{B}}$ here, according to question resultant FIELD is INCLINED at 45° with earth's field. so, $\theta$ =45° so, tan45°= 1 = $\frac{B_H}{B}$ so, $B_H=B=\frac{\mu_0}{4\pi}\frac{M}{r^3}$ Given, B =0.42 × 10^-4 T , M = 5.25 × 10^-2J/T so, 0.42 × 10^-4 = 10^-7 × 5.25 × 10^-2/r³ r³ = 12.5 × 10^-5 = 125 × 10^-6 r = 5 × 10^-2 m = 5cm (b) at axis of magnet, tan45° = 1 = B/ $B_H$ $B_H=B=\frac{\mu_0}{4\pi}\frac{2M}{r^3}$ so, 0.42 × 10^-4 = 10^-7 × 2 × 5.25 × 10^-2/r³ r³ = 25 × 10^-5 = 250 × 10^-6 r = 6.3 CM

Discussion

8.	A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the center, The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts?
Answer» *Given in the question :-* Rod MASS = m Rod length = l force acting perpendicular from the DISTANCE L/4 *Now the torque which acts on the rod* $T = F * \frac{L}{4}$ *Moment of INERTIA of the rod* $I= \frac{mL^2}{12}$ Angular ACCELERATION of the rod, α = T/I α = FL12/4mL²* $\alpha = \frac{3F}{mL}$ *At time t , angle rotation in the rod* $\theta = 0 * t+\frac{1}{2} at^2$ =1/2 × (3F/mL) × t² θ= 3Ft²/2mL *Hope it Helps :-)*

8.

A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the center, The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts?

Answer»

Given in the question :-

Rod MASS = m

Rod length = l

force acting perpendicular from the DISTANCE L/4

Now the torque which acts on the rod

$T = F * \frac{L}{4}$

Moment of INERTIA of the rod

$I= \frac{mL^2}{12}$

Angular ACCELERATION of the rod,

α = T/I

α = FL*12/4mL²

$\alpha = \frac{3F}{mL}$

At time t , angle rotation in the rod

$\theta = 0 * t+\frac{1}{2} at^2$

=1/2 × (3F/mL) × t²

θ= 3Ft²/2mL

Hope it Helps :-)

Discussion

9.	A pendulum is hanging from the ceiling of a cage T1, T2 are the tensions in the string. When the cage is moving upward and downward with same acceleration. What is the tension in the string if is move horizontaly? (solve in IIT method).
Answer» Final ANSWER : $T = \sqrt{\frac{T_1^2+T_2^2}{2}}$ Steps: 1) Let the acceleration be 'a' in all cases. We got, When acceleration is downward, $mg-T_2=ma \\ => T_2 =m(g-a) --(1)$ 2) When acceleration is upward , $T_1 -mg = ma \\ => T_1 = m(g+a) --(2)$ 3) When acceleration is horizontal, $T \sin(\theta) = ma \\ T \cos(\theta) = mg \\ => T^2 = (mg)^2 + (ma)^2 \\ => T = m \sqrt{g^2 + a^2 }$ 4) Squaring EQ. (1) and (2) : $T_1^2 + T_2^2 = 2m^2( a^2 + g^2) \\ => T_1^2+ T_2^2 = 2T^2 \\ => T = \sqrt{\frac{T_1^2+T_2^2}{2}}$ Hope , you got desired answer.

Discussion

10.	A body rotating at 20 rad /s is acted upon by a constant torque providing it a deceleration of 2 rad/s², At what time will the body have kinetic energy same as the initial value if the torque continues to act . ?
Answer» *Given in thew question :-* Initial velocity = 20 rad / second. $\alpha$ therefore = 20 rad/second² *Due to the TORQUE the effect of DECLARATION on the body , the body will start accelerating in OPPOSITE direction.* Therefore the initial Kinetic ENERGY = K.e at any instant. Hence Init. velocity = velocity at any instant $\omega'$ Now Fin. velocity =(-ω)= -20 rad/ second. $\omega = 20 rad/second$ Init. velocity *Therefore we know the formula ,* $\omega' = \omega - \alpha t$ $\omega' = -\omega$ since Hence, $-\omega = \omega - \alpha t$ 2.ω = α t t = (2 × ω) / α t = (2 × 20 )/2 *t = 20 second.* *Hope it Helps :-)*

10.

A body rotating at 20 rad /s is acted upon by a constant torque providing it a deceleration of 2 rad/s², At what time will the body have kinetic energy same as the initial value if the torque continues to act . ?

Answer»

Given in thew question :-

Initial velocity = 20 rad / second.

$\alpha$ therefore = 20 rad/second²

Due to the TORQUE the effect of DECLARATION on the body , the body will start accelerating in OPPOSITE direction.

Therefore the initial Kinetic ENERGY = K.e at any instant.

Hence Init. velocity = velocity at any instant

$\omega'$ Now Fin. velocity =(-ω)= -20 rad/ second.

$\omega = 20 rad/second$ Init. velocity

Therefore we know the formula ,

$\omega' = \omega - \alpha t$

$\omega' = -\omega$ since

Hence,

$-\omega = \omega - \alpha t$

2.ω = α t

t = (2 × ω) / α

t = (2 × 20 )/2

t = 20 second.

Hope it Helps :-)

Discussion

11.	at time a particle starts moving along the x-axis. if its kinetic energy increases uniformly with time ???t??? the net force acting on it must be proportional to:
Answer» It MUST be PROPORTIONAL to t

Discussion

12.	A man can swim in still water at a speed of 6kmph and he has to cross the river and reach just opposite point on the other bank. if the river is flowing at a speed of 3kmph, he has to swim in the direction
Answer» Final ANSWER : $tan^{-<klux>1</klux>}(1/2)$ with the vertical. Steps : 1) Man has to swim in such a way that, he WANT to reach just opposite point on the other bank. => Drift is 0 . 2)Speed of River is 3km/h in horizontal direction. => For Drift to be zero, Horizontal component of velocity of man should be 3km/h leftward. as SHOWN. We get, $\|v_x\| = 3 km/h \\ \|v_y\| = 6 km/h \\ => \tan(\theta) = \frac{\|v_x\|}{\|v_y\|} \\ => \tan(\theta) = \frac{1}{2}$ Hence, Man has to swim by making angle $\theta$ with vertical.

Discussion

13.	A block of mass 2 kg is placed on the floor . The coefficient of static friction between the two surfaces is 0.4. A horizontal force of 2.8 N is applied on the block parallel to the floor, what is the force of friction between the block and the floor? g=9.8m
Answer» The friction FORCE maximum=0.4(2)(9.8)=7.84N. the force acting on BLOCK is 2.8N. but friction force can't EXCEED the applied force,if so the block will MOVE in opposite DIRECTION of applied force. so friction force is 2.8N or -2.8N with sign.

Discussion

14.	two lens of +2.5 dioptre and 0.5 dioptre are placed in contact. find the power and focal length of lens combination. find focal length of lens combination also?
Answer» Here = P1 = 2.5 D , P2 = 0.5 D Power of Combination , P = P1 + P2 = 2.5 + 0.5 = 3.0 D Focal LENGTH of combination , F = 1/p = 3/1.0 = 30

14.

two lens of +2.5 dioptre and 0.5 dioptre are placed in contact. find the power and focal length of lens combination. find focal length of lens combination also?

Answer»

Here = P1 = 2.5 D , P2 = 0.5 D

Power of Combination ,

P = P1 + P2 = 2.5 + 0.5 = 3.0 D

Focal LENGTH of combination ,

F = 1/p = 3/1.0 = 30

Discussion

15.	A train starting from rest, pick up a speed of 10m/s in 100s. It continues to move at the same speed for the next 250s. It is then brought to rest in the next 50s. Plot a speed – time graph for the graph. Calculate a) the acceleration b) retardation c) total distance travelled by the train
Answer» Hey dear..... First let us plot the graph with the data given. Let the starting point be the ORIGIN in the speed - time graph. (a) Uniform rate of acceleration = slope of AB = (20−0) (m/s)/ [(200−0) s] = 0.1 m/s2. (b) Uniform retardation = − slope of CD = − (0−20)(m/s) /[(800−700)s] = 0.2 m/s2. (c) Total distance COVERED before stopping = area under the curve ABCD = area of ΔABE + area of rectangle BCFE + area of ΔCDF = ½ x 200 x 20 + 500 x 20 + ½ x 100 x 20 m = 2000 + 10000 + 1000 m = 13,000 m = 13 km (d) Average speed = Total distance covered/total time taken = 13000 m /(800 s) = 16.25 m/s HOPE IT HELPS ✌ ✌ ✌

Discussion

16.	A 10 kg ball is dropped from a height of 10m. Find(a) the initial potential energy of the ball , (b) the kinetic energy just before it reaches the ground, and (c) the speed just before it reaches the ground. (Ans: 980J, 980J, 14m/s)
Answer» (a). Mass of the Body = 10 kg. Height = 10 m. Acceleration DUE to gravity = 9.8 m/s². Using the Formula, Potential Energy = mgh = 10 × 9.8 × 10 = 980 J. (B). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant. ∴ Kinetic Energy before the body REACHES the ground is equal to the Potential Energy at the height of 10 m. ∴ Kinetic Energy = 980 J. (c). Kinetic Energy = 980 J. Mass of the ball = 10 kg. ∵ K.E. = 1/2 × mv² ∴ 980 = 1/2 × 10 × v² ∴ v² = 980/5 ⇒ v² = 196 ∴ v = 14 m/s. Hope it helps.

16.

A 10 kg ball is dropped from a height of 10m. Find(a) the initial potential energy of the ball , (b) the kinetic energy just before it reaches the ground, and (c) the speed just before it reaches the ground. (Ans: 980J, 980J, 14m/s)

Answer»

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration DUE to gravity = 9.8 m/s².

Using the Formula,

Potential Energy = mgh

= 10 × 9.8 × 10

= 980 J.

(B). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body REACHES the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

Hope it helps.

Discussion

17.	Where do you observe Pascal’s principle in our daily life? Give few examples.
Answer» Pascal's law is everywhere in our DAILY LIFE: examples are 1)hydraulic brakes 2)hydraulic bridges 3)hydraulic jacks 4)hydraulic pumps etcetera..... hope it helps U....

Discussion

18.	Two small glass spheres of mass 10g and 20g are moving in a straight line in the same direction with the velocities of 3m/s and 2m/s respectively. They colied with each other and after collision, glass sphere of mass 10g moves with a velocity of 2.5 m/s. Find the velocity of second ball after collision.ANSWER IT FAST EXTREMELY URGENT.
Answer» INITIAL momentum of first SPHERES = MV = 0.1×3=0.3kg m/s initial momentum of second SPHERE = mv = 0.2×2 = 0.4kg m/s After the collision final momentum of first sphere= mv = 0.1×2.5=0.25kg m/s final momentum of second sphere =mv =0.2×v = 0.2kg m/s initial momentum = final momentum 0.3+0.4 = 0.25 + 0.2v 0.7 = 0.25 +0.2v 0.45 = 0.2v v= 2.25m/s

Discussion

19.	When you lift a box from the floor and put it on an almirah the potential energy of the box increases but there is no change in its kinetic energy. Is it violation of conservation of energy? Explain.
Answer» Law of the Conservation of the energy states that the total sum of the kinetic and the potential energy remains the constant. Also, It states that the energy can neither be created nor be destroyed. It can only be converted into one form to another. No, It is not the VIOLATION of the Principle of the Conservation of the Energy. The Energy of the system is conversed when there is no EXTERNAL force applied on the body. In the Given case, External Force have DONE the work on the box to take it from floor to the ALMIRAH. By doing this, The potential energy of the box has been increased and the Kinetic Energy remains the constant. Hence, it is not the violation of the law of the conservation of the energy. Hope it helps you.

19.

When you lift a box from the floor and put it on an almirah the potential energy of the box increases but there is no change in its kinetic energy. Is it violation of conservation of energy? Explain.

Answer»

Law of the Conservation of the energy states that the total sum of the kinetic and the potential energy remains the constant.

Also, It states that the energy can neither be created nor be destroyed. It can only be converted into one form to another.

No, It is not the VIOLATION of the Principle of the Conservation of the Energy.

The Energy of the system is conversed when there is no EXTERNAL force applied on the body.

In the Given case, External Force have DONE the work on the box to take it from floor to the ALMIRAH. By doing this, The potential energy of the box has been increased and the Kinetic Energy remains the constant.

Hence, it is not the violation of the law of the conservation of the energy.

Hope it helps you.

Discussion

20.	Brainliest question do hurry i mark brainliest
Answer» HOPE it will HELP you PLZZZZ MARK me as BRAINLIST

Discussion

21.	a bomber plane moves horizontally with speed of 500m/s and a bomb released from it strikes the ground in 10 sec. angle at which it strikes the ground will be?? take g =10m\s2
Answer» It is RECTANGLE RULE of VECTOR addition and HOPE it helps you

Discussion

22.	What is electric current. Name and define its S I unit in a conductor electrons flowing from B to A . What is the direction of conventional current. Give justification for your answer. A steady current of 1 Ampere flows through a conductor.Calculate the number of electrons that flows through any section in 1 second? (charge of electrons 1.6 * 10^-9.....Plz Help...
Answer» Current is the NET flow of charged particles. It's measured in Amperes. Current is modeled as the flow of positively charged particles, so that if ELECTRONS flow from B to A, convention states that current FLOWS from A to B. This is the convention because Benjamin FRANKLIN decided it a few centuries ago. 1 A * 1 s = 1 C 1 C * 1 electron / (1.6 * 10^-19 C) = 6.25*10^18

22.

What is electric current. Name and define its S I unit in a conductor electrons flowing from B to A . What is the direction of conventional current. Give justification for your answer. A steady current of 1 Ampere flows through a conductor.Calculate the number of electrons that flows through any section in 1 second? (charge of electrons 1.6 * 10^-9.....Plz Help...

Answer»

Current is the NET flow of charged particles. It's measured in Amperes. Current is modeled as the flow of positively charged particles, so that if ELECTRONS flow from B to A, convention states that current FLOWS from A to B. This is the convention because Benjamin FRANKLIN decided it a few centuries ago.

1 A * 1 s = 1 C

1 C * 1 electron / (1.6 * 10^-19 C) = 6.25*10^18

Discussion

23.	During a heavy rain, hailstones of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstones strike every square meter of a 10m x 10m roof perpendicularly in one second and assume that the hailstones do not rebound. Calculate the average force exerted by the falling hailstones on the roof. Density of a hailstone is 900 kg/m³.
Answer» Thanks for asking the QUESTION! ANSWER:: Diameter of hailstone , d = 1.0 cm v = 20 m/s u = 0 ρ = 900 kg/m³ = 0.9 gm/cm³ Volume of hailstone = (4/3)πr³ = (4/3)π(0.5)³ = 0.5238 cm³ Mass of hailstone = vρ = 0.5238 x 0.9 = 0.4714258 gm Mass of 2000 hailstone = 2000 x 0.4714258 = 942.8516 Rate of change in MOMENTUM per unit area = 942.8516 x 2000 = 1885703.2 =18.85 kg m/s² = 18.85 N Total force on roof = 18.85 x 10 x 10 = 1885 N ≈ 1900 N Hope it helps!

23.

During a heavy rain, hailstones of average size 1.0 cm in diameter fall with an average speed of 20 m/s. Suppose 2000 hailstones strike every square meter of a 10m x 10m roof perpendicularly in one second and assume that the hailstones do not rebound. Calculate the average force exerted by the falling hailstones on the roof. Density of a hailstone is 900 kg/m³.

Answer»

Thanks for asking the QUESTION!

ANSWER::

Diameter of hailstone , d = 1.0 cm

v = 20 m/s

u = 0

ρ = 900 kg/m³ = 0.9 gm/cm³

Volume of hailstone = (4/3)πr³ = (4/3)π(0.5)³ = 0.5238 cm³

Mass of hailstone = vρ = 0.5238 x 0.9 = 0.4714258 gm

Mass of 2000 hailstone = 2000 x 0.4714258 = 942.8516

Rate of change in MOMENTUM per unit area = 942.8516 x 2000 = 1885703.2

=18.85 kg m/s²

= 18.85 N

Total force on roof = 18.85 x 10 x 10 = 1885 N ≈ 1900 N

Hope it helps!

Discussion

24.	A few drops of ethanoic acid were added to solid sodium carbonate. The possible results of thereactions are: [ ]a) A hissing sound was evolvedb) Brown fumes evolved.c) Brisk effervescence occurred. d) A pungent smelling gas evolved.
Answer» Yes. The ACID used or sodium CARBONATE used should be ENOUGH to PRODUCE the sound which is due to the evolution of CARBONDIOXIDE. 2CH3COOH + Na2CO3 --> 2CH3COONa + H2O + CO2 Option A is correct

Discussion

25.	What should be the position of the object when a concave mirror is to be used as a shaving mirror
Answer» 1) A concave mirror produces a virtual and magnified IMAGE when the objectis PLACED very close (between P and F, or between C and F) to it. This position should be USED for shaving mirror. 2) When the object is placed at the FOCUS, it generates a parallel beam of light. So for torches the object should be at F.

Discussion

26.	A 0.2 kg object at rest is subjected to a force 0.3i-0.4j Newton what will be velocity after 6 sec
Answer» Force = MASS × Acceleration Given that the force is in vector form we can write it in form of COLUMN vector. After this we can substitute the values of force and mass to get acceleration from this. Knowing that acceleration = change in VELOCITY ÷ time We can then get the final velocity from this formula. The velocity and the acceleration will be in vector form. Find the calculations in the images below. From the images we can see that the the final velocity is : 9I - 12j

Discussion

27.	If the unit of force and length each be increased by four then the unit of work is increased by
Answer» 16 times W = FD W' = 4F4D = 16FD = 16W

27.

If the unit of force and length each be increased by four then the unit of work is increased by

Answer»

16 times

W = F*D

W' = 4F*4D = 16FD = 16W

Discussion

28.	Why is it easier to carry the same amount of water in two buckets, one in each hand rather than in a single bucket?
Answer» THANKS for the question! It is definitely a very interesting question to solve and do some brainstorming. ************************************************ It is easier to carry two buckets of water in one hand each, than to carry only one in one hand becausethe centre of gravity and centre of EQUILLIBRIUM fall within the FEET. The weight is SPACED out more evenly on both the sides of the body than it would be if you had to carry in a single BUCKET. ************************************************ Hope it helps and solves your query!!

28.

Why is it easier to carry the same amount of water in two buckets, one in each hand rather than in a single bucket?

Answer»

THANKS for the question!

It is definitely a very interesting question to solve and do some brainstorming.

**************************************************

It is easier to carry two buckets of water in one hand each, than to carry only one in one hand becausethe centre of gravity and centre of EQUILLIBRIUM fall within the FEET.

The weight is SPACED out more evenly on both the sides of the body than it would be if you had to carry in a single BUCKET.

**************************************************

Hope it helps and solves your query!!

Discussion

29.	a bullet of mass 0.015 kg travels with horizontal speed 50 ms?? strikes a block of wood of mass 0.2 kg and instantly comes to rest with respect to the block. the block is suspended from the ceiling by means of thin wire. calculate the height to which the block rises.
Answer» Is that ENOUGH???? OKAY

Discussion

30.	A needle 4 cm high is placed in front of a lens. Image of the needle is real, inverted and 6 cm high, when the distance between the needle and its image is 20 cm. find the focal length of lens and also write the type of the lens.
Answer» +20 is the FOCAL LENGTH of the LENS.focal is +ve so it,s a CONVEX lens.

Discussion

31.	A man carrying a bag of total mass 25kg climbs up to a height of 10m in 50 seconds. Calculate the power delivered by him on the bag. ( Ans : 49J)
Answer» Given conditions ⇒ Mass of the BODY = 25 KG. Height = 10 m. Time = 50 seconds. Kinetic Energy = mgh = 25 × 9.8 × 10 = 2450 J Now, ∵ POWER = Energy/Time ∴ P = 2450/50 ∴ P = 49 W. Hence, the Power used by the man is 49 W. Hope it helps.

31.

A man carrying a bag of total mass 25kg climbs up to a height of 10m in 50 seconds. Calculate the power delivered by him on the bag. ( Ans : 49J)

Answer»

Given conditions ⇒

Mass of the BODY = 25 KG.

Height = 10 m.

Time = 50 seconds.

Kinetic Energy = mgh

= 25 × 9.8 × 10

= 2450 J

Now,

∵ POWER = Energy/Time

∴ P = 2450/50

∴ P = 49 W.

Hence, the Power used by the man is 49 W.

Hope it helps.

Discussion

32.	Question 7.3: A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.Class 12 - Physics - Alternating Current Alternating Current Page-266
Answer» RMS CURRENT is given by $I_V=\frac{E_v}{X_L}$ here, $I_v$ is the RMS VALUE of current in CIRCUIT. $E_v$ is the voltage supply by BATTERY e.g., $E_v$ =220V and $X_L$ is the resistance of inductor, e.g., $X_L=2\pi\nu L$ here v is the frequency and L is inductance of inductor.e.g., XL = 2π × 50 × 44 × 10^-3 ohm now, Iv = 220/(2π × 50 × 44 × 10^-3) = 15.9A

Discussion

33.	Applications of magnetic effect of electric current in daily life
Answer» The generators, motors, loudspeakers, HEADPHONES are some applications of magnetic EFFECTS of ELECTRIC CURRENT in daily life .. hope it HELPS you!!

Discussion

34.	from the top of a tower of height 40m a ball is projected upward with a speedof 20m/sec at an elevation of 30degrees . the ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the point of projection is g=10m/s
Answer» Time TAKEN to reach maximum height in a horizontal projection : H = 20²sin²30° / 2 × 10 H = (400 × 0.25) / 20 = 5m TOTAL height is : 5 + 40 = 45 m Time of flight = (2 × 20 × SIN 30) / 10 = 2 seconds Time of FREE fall is √(2 × 45)/10 = √9 = 3 seconds. The total time taken to reach the ground is thus : 1 + 3 = 4 Seconds The ratio is thus : 4 : 2

Discussion

35.	containers of same volume with atomic hydrogen and Helium respectively at 1 and 2 ATM pressure if the temperature of both specimen are same then average speed of for hydrogen atoms will be
Answer» HEYA MATE, HERE IS UR ANSWER. Thanks for ASKING this question. You said equal mass of Helium, since the MOLECULAR weight of He is double that of H2, this means you have twice as many H2 molecules as Helium. Everything ELSE being equal, HALF of the molecules will then give you half of the pressure in this case 1 atm of Helium. The trap here it that you base on weight instead of molecules. If the number of molecules were the same, then the pressure WOULD be the same too but the mass of He would then be double. I HOPE IT HELPS U.

Discussion

36.	Can u solve this question if yes
Answer» USE this SHORT cut to GET answer quickly

Discussion

37.	When tall buildings are constructed on earth, the duration of day-night slightly increases. Is it true?
Answer» Answer ⇒ Yes, the DURATION of day-night slightly increases when tall buildings are constructed on earth. Explanation ⇒ When the tall buildings are constructed, the EFFECTIVE radius of the earth slightly (means very small, approximately zero. Sometimes EXPRESSED as the Limit tends to zero) increases. We know that the mass remains the constant, thus the moment of inertia of the earth which is represented by 'l' is slightly increases. To conserve the angular momentum ( represented by L) of the earth the angular velocity (ω = L/I) will slightly decreases. This means that the earth rotates slower now and takes more time to complete one ROTATION. Thus the the duration of day-night slightly increases. Hope it helps.

37.

When tall buildings are constructed on earth, the duration of day-night slightly increases. Is it true?

Answer»

Answer ⇒ Yes, the DURATION of day-night slightly increases when tall buildings are constructed on earth.

Explanation ⇒ When the tall buildings are constructed, the EFFECTIVE radius of the earth slightly (means very small, approximately zero. Sometimes EXPRESSED as the Limit tends to zero) increases.

We know that the mass remains the constant, thus the moment of inertia of the earth which is represented by 'l' is slightly increases.

To conserve the angular momentum ( represented by L) of the earth the angular velocity (ω = L/I) will slightly decreases. This means that the earth rotates slower now and takes more time to complete one ROTATION. Thus the the duration of day-night slightly increases.

Hope it helps.

Discussion

38.	a vehicle travels half the distance l with speed v1 and the other half with speed v2 then its average speed is - solution
Answer» FORMULA for AVERAGE speed=total distance/total time Let distance traveled =D time taken in completing first HALF distance =D/2(V1). time taken to complete SECOND half distance =D/2v2. so total time taken =D/2v2 + D/2v1 =D(v1+v2)/2v1v2. so average speed=D/[D(v1+v2)/2v1v2] =2v1v2/(v1+v2).

Discussion

39.	What is meant by uniform circular motion
Answer» When a body is COVERING equal DISTANCE which equal acceleration on a circular PATH in equal TIME INTERVAL then it is known as uniform circular motion

Discussion

40.	b) intensity of light coming out of a polaroid does not change irrespective of the orientation of the pass axis of the polaroid. explain why.
Answer» MONOCHROMATIC LIGHT is STABLE

Discussion

41.	If r the radius and a is th mass number of a necleus then the graph of log r versus log a will be
Answer» #1 : We KNOW that Relation : $<klux>R</klux> = R_o A^{1/3} --(1)$ where R = NUCLEAR Radius $R_o = 1.2 * 10^{-15} =1.2 fm$ A = Mass Number #2 : We have, Taking log in both sides of EQUATION (1) : $log(R) = log(R_o) + \frac{1}{3} log(A) ----(1)$ Hence, Graph of log(R) VS. log(A) will be a straight line whose : Slope : 1/3 y- intercept = $log(R_o)$

Discussion

42.	a person walks a distance of 30 m towards west with the speed of 2m/s and 40m towards north 1.5m/s. find average speed and average velocity
Answer» This is your ANSWER HOPE you UNDERSTAND PLEASE THANK

Discussion

43.	Ues. a car is 250 metres behind the bus. the car and bus are moving with speed 60 km/hr and 35 km/hr respectively. the car will be ahead of bus by 250 metres in:
Answer» In this process the CAR have to cover 250+250=500metres relative to bus. relative VELOCITY of car wrt bus=60-35=25m/s. so time taken to COMPLETE 500M relative distance with 25m/s relative velocity =500/25=20sec

Discussion

44.	Definite value of energy possessed by a quantum of radition is called
Answer» The VALUE of energy possessed by a QUANTUM of RADIATION is CALLED photon.

Discussion

45.	A particle is projected with a velocity 49 m/s at an angle of 30 degree to the horizontal. calculate the maximum height, time of flight and horizontal range?
Answer» PLEASE MARK it as BRAINLIEST

Discussion

46.	rain is falling vertically downwards with a velocity of 3 km/hr. a man walks in the rain with a velocity of 4 km/hr. the raindrops will fall on the man with a velocity of
Answer» USE HYPOTENUSE THEOREM

Discussion

47.	As a particle moves from the origin to left can the displacement be positive
Answer» No when the DISPLACEMENT moves from ORGIN to left then the displacement is NEGATIVE and it is KNOWN as retardution

Discussion

48.	How to do this sum? converting 1g/cm3= 1g---- 1cm3what is the answer?
Answer» What ......... 1g/cm3 to.....

Discussion

49.	What is the value of linear velocity if w=3i-4j+5k and r=5i-6j+6k?
Answer» USE v=rw Hope it will HELP

Discussion

50.	A person walks along a circular path of radius r from a to b as shown in the figure . displacement transversed by the person as
Answer» The ANSWER for the QUESTION is 5/6Лr MARK me as BRAINLIEST

Discussion

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