Physics
General
Easy

Question

The maximum speed with which a car can cross a convex bridge over a river with radius of curvature 9 m is : (given that the centre of gravity of car is 1m above the road)-

  1. 50 m/s  
  2. 30 m/s  
  3. 20 m/s  
  4. 10 m/s  

The correct answer is: 10 m/s

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Physics-

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K subscript 1 end subscript comma K subscript 2 end subscript blank a n d blank K subscript 3 end subscript as shown. If a single dieletric material is to be used to have the same capacitance C is this capacitors, then its dielectric constant K is given by

Capacitance of two capacitors each of area fraction numerator A over denominator 2 end fraction blank comma plate separation d but dielectric constants K subscript 1 end subscript and K subscript 2 end subscriptrespectively joined in parallel
C subscript 1 end subscript equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction equals fraction numerator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A over denominator d end fraction
It is in series with a capacitor of plate area A comma plate separation d divided by 2 blankand dielectric constantK subscript 3 end subscript blank i e comma blank C subscript 2 end subscript equals fraction numerator K subscript 3 end subscript epsilon subscript 0 end subscript A over denominator d divided by 2 end fraction.
If resultant capacitance be taken as C equals fraction numerator K subscript blank epsilon subscript 0 end subscript A over denominator d end fraction,
Then fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C subscript 1 end subscript end fraction plus fraction numerator 1 over denominator C subscript 2 end subscript end fraction
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rightwards double arrow fraction numerator 1 over denominator K end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript plus blank K subscript 2 end subscript end fraction plus fraction numerator 1 over denominator 2 K subscript 3 end subscript end fraction

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Capacitance of two capacitors each of area fraction numerator A over denominator 2 end fraction blank comma plate separation d but dielectric constants K subscript 1 end subscript and K subscript 2 end subscriptrespectively joined in parallel
C subscript 1 end subscript equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d divided by 2 end fraction equals fraction numerator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A over denominator d end fraction
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If resultant capacitance be taken as C equals fraction numerator K subscript blank epsilon subscript 0 end subscript A over denominator d end fraction,
Then fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C subscript 1 end subscript end fraction plus fraction numerator 1 over denominator C subscript 2 end subscript end fraction
therefore fraction numerator d over denominator K epsilon subscript 0 end subscript A end fraction equals fraction numerator d over denominator open parentheses K subscript 1 end subscript plus K subscript 2 end subscript close parentheses epsilon subscript 0 end subscript A end fraction plus fraction numerator d divided by 2 over denominator table row cell K subscript 3 end subscript epsilon subscript 0 end subscript A end cell row cell end cell end table end fraction
rightwards double arrow fraction numerator 1 over denominator K end fraction equals fraction numerator 1 over denominator K subscript 1 end subscript plus blank K subscript 2 end subscript end fraction plus fraction numerator 1 over denominator 2 K subscript 3 end subscript end fraction
General
Physics-

A body weight 8 g when placed in one pan and 18 g when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty. The true weight of the body is

Let the mass of first and second pan be m1 and m2, respectively and their respective distance from the pivot be L1 and L2.

Let the true weight of the body be M.

As the beam is horizontal when both the pans are empty, thus net torque about the pivot is zero.
  m1gL1m2gL2=0
We get  m1L1=m2L2      ....(1)
The body weighs 8g in first pan.
  (M+m1)gL1=(8+m2)gL2
Or   ML1=8L2    ....(2)        (Using (1))
The body weighs 18g in second pan.
  (18+m1)gL1=(M+m2)gL2
Or   18L1=ML2    ....(3)        (Using (1))
Dividing equation (2) and (3) we get  18M=M8
Or  M2=8×18=144
  M=12 g 

A body weight 8 g when placed in one pan and 18 g when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty. The true weight of the body is

Physics-General
Let the mass of first and second pan be m1 and m2, respectively and their respective distance from the pivot be L1 and L2.

Let the true weight of the body be M.

As the beam is horizontal when both the pans are empty, thus net torque about the pivot is zero.
  m1gL1m2gL2=0
We get  m1L1=m2L2      ....(1)
The body weighs 8g in first pan.
  (M+m1)gL1=(8+m2)gL2
Or   ML1=8L2    ....(2)        (Using (1))
The body weighs 18g in second pan.
  (18+m1)gL1=(M+m2)gL2
Or   18L1=ML2    ....(3)        (Using (1))
Dividing equation (2) and (3) we get  18M=M8
Or  M2=8×18=144
  M=12 g