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Lesson Posted on 08 Jan CBSE/Class 11/Science/Physics

Tushar Lohani

I am perusing my B.tech from SRM University in computer science and I have an experience of 2 years in...

Physics is not only a subject it is an observation. So if someone wants to learn physics first thing he needs to learn is just how to observe things. Just take an example of a guy who is sitting inside a train, and you are an observer who is observing that guy from the bench which is situated on the... read more

Physics is not only a subject it is an observation. So if someone wants to learn physics first thing he needs to learn is just how to observe things. Just take an example of a guy who is sitting inside a train, and you are an observer who is observing that guy from the bench which is situated on the platform so what you say that guy is moving, or he is in rest. So your answer is just that he is running with the exact velocity from which the train is moving. Now assume that you are sitting inside the train then what you will say that guy in rest because you are also running with the same velocity from which your observer moves, so that's why I told physics need observation.

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Answered on 05 Jan CBSE/Class 11/Science/Physics

Nizamuddin khan

Tutor

The current flowing through a object is directly proportional to the potential difference applied across the object. i is proportional to V and 1/R is the proportional constant.
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Education

Due to reflection of sun
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Mohit Singla

No
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The efficiency of a Carnot cycle is . If on reducing the temperature of the sink by 65°C, the efficiency... read more
The efficiency of a Carnot cycle is $\frac{1}{6}$. If on reducing the temperature of the sink by 65°C, the efficiency becomes $\frac{1}{3}$  , determine the initial and final temperatures 3 between which the cycle is working. read less

Spandan Maikap

Efficiency of a Carnot cycle,=1-(Tsink/Tsource) If eff =1/6 then Tsink/Tsource= 5/6 ------(1) If eff = 1/3 the Tsink-65/Tsource= 2/3------(2) Solving the equation 1 and 2 T sink = 325( lower temp) T source = 390( higher temperature) read more

Efficiency of a Carnot cycle,=1-(Tsink/Tsource)

If eff =1/6 then Tsink/Tsource= 5/6 ------(1)

If eff = 1/3 the Tsink-65/Tsource= 2/3------(2)

Solving the equation 1 and 2

T sink = 325( lower temp)

T source = 390( higher temperature)

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Answered on 03 Jan CBSE/Class 11/Science/Physics

State Archimedes' principle.

Romy Gupta

It states that whenever a body is partially or fully immersed in a liquid the body will experience an upthrust(upward force). This upward force will be equal to the weight of the liquid displaced by the body.
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If a solid body floating in water has th of the volume above surface, then determine the fraction of... read more
If a solid body floating in water has $\frac{1}{6}$th of the volume above surface, then determine the fraction of its volume that will project upwards if it floats in a liquid of specific gravity 1.2 read less

Spandan Maikap

Concepts involved : 1) specific gravity = density of liquid / density of water 2) boyancy force = volume immersed *density of liquid (ρ)* g( accn due to gravity) 3) weight of the object = total volume *density (σ)*g Now since 1/6 th of the volume is above liquid surface 5/6 th is below it... read more

Concepts involved :

1) specific gravity = density of liquid / density of water

2) boyancy force = volume immersed *density of liquid (ρ)* g( accn due to gravity)

3) weight of the object = total volume *density (σ)*g

Now since 1/6 th of the volume is above liquid surface 5/6 th is below it therefore by force balancing ( the upward boyancy force and downward gravity )

in case of water  5/6*V*ρ*g = V*σ*g ----(1)

in case of the other liquid  λ*V*ρ'*g=V*σ*g ----(2)

Therefore from 1 and 2

λ*ρ' =5/6*ρ

λ*1.2=5/6*1

λ=25/36

Therefore the fraction of thw volume outside the  surface of water = 9/36

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A body of mass m moving with an initial velocity u collides inelastically with a body of mass M initially... read more
A body of mass m moving with an initial velocity u collides inelastically with a body of mass M initially at rest. If the collision is completely inelastic, then find (i) final velocity of combined system. (ii) loss in kinetic energy during collision. read less

Ajay

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass 'm+M' move with final velocity let us say 'v' which we have to find out If the external force on system is zero then by law of conservation of momentum of system , we have , initial... read more

(i) As the collision is completely inelastic therefore the two bodies stick together and the sticked bodies of mass  'm+M' move with final velocity let us say 'v'  which we have to find out

If the external force on system is zero then by law of conservation of momentum of system , we have ,

initial momentum of system = final momentum of system

⇒mu +M×0 = (m+M) v

⇒(m+M) v = mu

⇒v = mu ⁄ (m+M)

Thus the above value of v is the required final velocity

(ii)  initial kinetic energy of system =

½mu² +½ M×0² = ½mu²

Final kinetic energy of system  = ½(m+M) v²

Now putting the value of v= mu⁄(m+M) in final KE

∴final KE  =½(m+M) × m² u²⁄ (m+M) ²

=½m² u² ⁄ (m+M)

∴ loss of kinetic energy = initial KE – final KE

=½mu² – ½ m²u² ⁄(m+M)

= {½mu²×(m+M) –½m²u²} ⁄ (m+M)

={½m²u² +½mMu² – ½m²u²}  ⁄ (m+M)

=½mMu² ⁄ (m+M)

∴loss of kinetic energy =½mMu² ⁄ (m+M)

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Vivek R

Teacher

Consider a 200kg bike was stand on ground.The action force 2000 N (W(m*g) = 200*10) is from bike towards center of Earth. The reactional force Fn = 2000 N was exerted from Earth to against the action force.. Both are not cancelled each other becoz the source of force from different obeject... read more

Consider a 200kg bike was stand on ground.The action force 2000 N (W(m*g) = 200*10) is from bike towards center of Earth. The reactional force Fn = 2000 N was exerted from Earth to against the action force.. Both are not cancelled each other becoz the source of force from different obeject...

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Ankush

youngs modules of perfectly rigid body is Zero
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