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JEE Advanced Physics Test 16

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JEE Advanced Physics Test 16
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  • Question 1
    4 / -1

    In the circuit shown (R = 1 Ω) current in branch CD is

    Solution

    Let us split the circuit and then rearrange it as shown:  

    Since this represents a balanced wheatstone bridge, current in CD branch is zero.  

     

  • Question 2
    4 / -1

    A Π s haped wire frame with a light, smooth, sliding wire connected to its open end (width of open end is l) is dipped in a soap solution and then raised, such that a soap film forms between the slider and the closed end of the wire frame. If the wire frame is now placed in a vertical plane and a block of mass m is hung from the slider then for mass m to remain in equilibrium the surface tension of the soap film is

    Solution

    Force due to surface tension,

    F = 2T × l  (since there are two free surfaces)

    For equilibrium F = mg

    ⇒  T = mg/2l

     

  • Question 3
    4 / -1

    An artificial satellite of mass m is moving in a circular orbit at a height equal to the radius R of the Earth. Suddenly due to internal explosion the satellite breaks into two parts of equal masses. One part of the satellite stops just after the explosion and then falls to the surface of the Earth. The increase in the mechanical energy of the system (satellite + Earth) due to explosion will be (take acceleration due to gravity on the surface of Earth as g) 

    Solution

     

  • Question 4
    4 / -1

    In position A kinetic energy of a particle is 60 J and potential energy is −20 J. In position B, kinetic energy is 100 J and potential energy is 40 J. Then in moving the particle from A to B

    Solution

    Work done by conservative forces  

    = Ui − Uf = −20 − 40 = −60 J

    Work done by external forces

    =  Ef – Ei = (Kf + Uf) – (Ki + Ui) = 40 – 140 =  −100 J  

    Work done by all the forces    

    = Kf – Ki = 100 – 60 = 40 J  

     

  • Question 5
    4 / -1

    A metal rod is fixed in horizontal position and a force of magnitude F is applied as shown. If RA = force by wall A and RB = force by wall B, then

    Solution

     

  • Question 6
    4 / -1

    Two waves travelling in opposite directions produce a standing wave. The individual wave functions are given by y1 = 4 sin (3x – 2t) cm and y2 = 4 sin (3x + 2t) cm, where x and y are in centimeter. Now, select the correct statement(s) from the following.

    Solution

    y = y1 +  y2  = (8 sin 3x. cos 2t) = Axcos 2t  

    Where Ax = 8  sin 3x

    ∴  Nodes are formed where 3x = 0, π, 2π, ….. etc.  

     

  • Question 7
    4 / -1

    A point source S is placed anywhere in between two converging mirrors having focal lengths f and 2f, respectively as shown. The value of d for which only single image may be formed is /are 

    Solution

    Only one image will be formed if the object is either at the common focus or common centre of curvature of both the mirrors. The ray diagrams are as shown. 

     

  • Question 8
    4 / -1

    Directions For Questions

    In the circuit shown, initially keys K1, K2 and K3 are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K1, K2 and K3.  

    Event 1: Keys K1 and K2 are closed and K3 is left open.  

    Event 2: After a long time K2 is opened and K3 is closed.  

    Event 3: After a long time K1 is opened and K2 is closed. 

    ...view full instructions

    The time constants for charging of capacitor (completion of Event 1) and rise of current through the inductor (completion of Event 2) respectively are  

    Solution

     

  • Question 9
    4 / -1

    Directions For Questions

    In the circuit shown, initially keys K1, K2 and K3 are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K1, K2 and K3.  

    Event 1: Keys K1 and K2 are closed and K3 is left open.  

    Event 2: After a long time K2 is opened and K3 is closed.  

    Event 3: After a long time K1 is opened and K2 is closed. 

    ...view full instructions

    Find the maximum charge that can come on the capacitor after the completion of all the three events 

    Solution


    Thus the capacitor again starts charging.  

    Charge on the capacitor will reach maximum value when current through the inductor becomes zero.  

    From conservation of energy between initial state (when i = 5 A, q = 40 μC) and final state (when i = 0, q = Qmax) we get

     

  • Question 10
    4 / -1

    Directions For Questions

    In the circuit shown, initially keys K1, K2 and K3 are all open. Now certain events as described below are performed successively on the circuit which involve closing and opening of the keys K1, K2 and K3.  

    Event 1: Keys K1 and K2 are closed and K3 is left open.  

    Event 2: After a long time K2 is opened and K3 is closed.  

    Event 3: After a long time K1 is opened and K2 is closed. 

    ...view full instructions

    Find the maximum current that can pass through the inductor after the completion of all the three events

    Solution

    After completion of Event 3

    The capacitor again starts charging, the current through inductor starts decreasing, becomes zero, changes direction and again starts increasing and goes beyond the value of 5 A.

    Current through the inductor will reach maximum value when charge on the capacitor has become zero.

    From conservation of energy between initial state (when i = 5 A, q = 40 μC) and final state (when i = Imax, q = 0) we get

     

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