2019-03-01_Physics_Times

2

4

(300)

9

T

 

  

 

 T 2 675 K

Rise in temperature = 675 - 300= 375 K

According to definition of cross-product

C D

 

is perpendicular to both A



and B.



i.e., A C  ( ) 0D

  

or A C   A D 0

   

or A(component of C



along A



) + A (component

of D



along A)



= 0

Heat received (or produced by the burning of

petrol) in hour will be

(2.4 / )(35.5 / )kg hour MJ kg

 85.2 10 /^6 J hour

 The rate at which heat is received

6

85.2 10 2.37 10 / 23.7 4

(3600 )

J

J s kW

s



   

The rate of heat rejection = rate at which heat is

produced- rate at which work is done

= 23.7 kW-10kW =13.7kW

   S S 1 S 2

To melt ice + to rise the temperature

2

1

T

T

Q mcdT

T T



 ^2

1

T

T

mL T

mc

T T



  

1 80 313

1 1 2.303log

273 273



   

(^)  0.28 0.1366 0.42 cal C ^1
Capacitance of a cylindrical capacitor
2
ln( / )
L
b a
 
Energy stored in the capacitor
1 1 ln( / )^222
[ ]
2 2 2
Q Q b a Q
U const
C L L
  

After doubling the charge and length U 2  2 U 1
Velocity of the particle is
4cos cos 40 .40 
3
dy x
t
dt

 
 
  
 
here
1
3,
8
x t
p 160 /
dy
v cm s
dt
   
Let T 1 and T 2 are tensions in the left and right
strings.
According to question,
(^1112)
2
T T
l l

 
T2 1T/
For rotational equilibrium,
T x T L x 1     2   x L/
Fundamental frequency of wire  
wire 2
v
f
l

(A)
3 5
, ,
4 4 4
v v v
f
l l l
 cannot match with
fwire
(B)
16.Sol:
17.Sol:
18.Sol:
19.Sol:
20.Sol:
21.Sol:
22.Sol:
23.Sol:
24.Sol:
2 3
, ,
2(2 ) 2(2 ) 2(2 )
v v v
f
l l l

Second harmonic
2
2(2 )
v
l
matches with fwire.
Similarly other pipes frequencies will not match with
fundamental frequency
Apparent frequency is equal to the frequency
since relative velocity is zero. So number of beats
will be zero.
The situation is shown in the fig. Both the source
(engine) and the observer (Person in the middle of
the train) have the same speed, but their direction
of motion is right angles to each other. The
component of velocity of observer towards source
is vcos45 and that of source along the line
joining the observer and source is also vcos45 .
There is no relative motion between them, so there
is no change in frequency heard. So frequency
heard is 200Hz.