Please wait...

Physics Test - 32
Result
Physics Test - 32
  • /

    Score
  • -

    Rank
Time Taken: -
  • Question 1/10
    1 / -0

    Four metallic plates each with a surface area of one side A, are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B as shown in the figure. Then the capacitance of the system is :

    Solutions

    There are two capacitors with capacitance C1=C2=Aϵ0d
    As they are in parallel so the net capacitance C=C1+C2=Aϵ0d+Aϵ0d=2Aϵ0d
    Hence, the correct option is (B)
  • Question 2/10
    1 / -0

    n equal capacitors are first connected in series and then in parallel. The ratio of capacitances in series and parallel arrangements will be :

    Solutions
    When n capacitors are connected in series, the equivalent capacitance is Cseries=Cn and when n capacitors are connected in parallel, the equivalent capacitance is Cparallel =nC CseriesCparallel=CnnC=1n2
    Hence, the correct option is (D)
  • Question 3/10
    1 / -0

    The minimum number of capacitors each of 3 μF required to make a circuit with an equivalent capacitance 2.25 μF is :

    Solutions
    There are four capacitors in which 3 capacitors are parallel and in series with 1 capacitor. Net Capacitance for 3 parallel capacitors =9μF Equivalent Capacitance for 3 parallel capacitors and in series with 1 capacitor =9×39+3=2.25μF
    Hence, the correct option is (B)
  • Question 4/10
    1 / -0

    When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with its combination. The resultant capacity is 3.75μF. The capacity of each capacitor is

    Solutions
    Let the capacitance be C when connected parallel, C1=C+C+C=3C when other was connected in series So 1Cf=13.75=13C+1C
    C=5μF
    Hence, the correct option is (A)
  • Question 5/10
    1 / -0

    A spherical drop of capacitance 1 μF is broken into 8 drops of equal radius. Then, the capacitance of each small drop is:

    Solutions
    Potential of a charged sphere is given by the formula
    V=kQR
    Hence, by Q=CV,
    C=QV=Rk
    As per given information, Rk=1μF
    After splitting,
    8(43πr3)=43πR3
    r=R2
    Thus, capacitance of each small drop is V=rk=12Rk=12μF
    Hence, the correct option is (A)
  • Question 6/10
    1 / -0

    Three capacitors of capacitance 1.0,2.0 and 5.0μF are connected in series to a 10V source. The potential difference across the 2.0μF capacitor is

    Solutions
    Ceq=1017μF,Q=10017μC
    Potential difference across 2μF capacitor =10017μC2μF=5017V
    Hence, the correct option is (C)
  • Question 7/10
    1 / -0

    A person has only two capacitors. By using them singly, in parallel or in series, he is able to obtain the capacitance 2μF,3μF,6μF and 9μF. What are the capacitance of the capacitors?

    Solutions
    For (A),Ceqs=2×32+3=65μF
    Ceqp=2+3=5μF
    For (B),Ceqs=3×63+6=2μF
    Ceqp=3+6=9μF
    For (C), Ceqs=6×96+9=185μF
    Ceqp=6+9=15μF
    For (D), Ceqs=2×92+9=1811μF
    Ceqp=2+9=11μF
    Thus only (B) will give the capacitances which are expectable.
    Hence, the correct option is (B)
  • Question 8/10
    1 / -0

    Two condensers each of capacitance 2μF are connected in parallel and this combination is connected in series with a 12μF capacitor. The resultant capacity of the system will be:

    Solutions
    When two capacitors of capacitance 2μF are in parallel, the equivalent capacitance is Ceq=2+2=4μF. Now it is connected to 12μF in series. So the capacitance of system is Csys=4×124+12=3μF
    Hence, the correct option is (D)
  • Question 9/10
    1 / -0

    A charge Q is divided into two parts. The two charges kept at a distance apart have a maximum columbian repulsion. Then the ratio of Q and one of the parts is given by :

    Solutions
    Total charge is Q Now let it is divided into q and (Q-q) Force of repulsion is F=q(Qq)r2
    For F to be maximum dFdq=0 Qqr2qr2=0
    Qq=2
    Hence, the correct option is (C)
  • Question 10/10
    1 / -0

    Two equally charged identical metal spheres A and B repel each other with a force F. Another identical uncharged sphere C is touched to A and then placed midway between A and B. The net force on C is in the direction:

    Solutions
    The spheres are equally charged. Let the charge be q. Now, using coulombs law, kq2/r2=F(r is the distance between the two spheres ) When the uncharged sphere touches A,q/2 charge gets transferred to C Now, force on C =k(q/2)(q/2)(r/2)2k(q/2)(q)(r/2)2=kq2/r22kq2/r2=F
    Hence, the correct option is (A)
User Profile
-

Correct (-)

Wrong (-)

Skipped (-)


  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Click on Allow to receive notifications
×
Open Now