575 - Physics Questions Answers
in first firing
m200 = 49mv so velocity of gun = 200/49
now apply momentum conservation again by taking 200/49 as the velocity of gun
The bob of mass m , charge q is circulating in a vertical circle of radius R. With the help of a string . If the maximum speed of the bob is V then the period of revolution is T1. If an electric field of magnitude (mg/q) is setup which makes an angle 600 with upward vertical . again the bob is circulating in same circle and its maximum speed is also v , then the period of revolution is T2. Find then T1/T2.
in first case net force of gravity = mg
in second case electric force = qE = qmg/q = mg
so resultant of gravitational mg and electric mg will be mg again at an angle 60 degree with vertically downward direction.
in this time max velocity will not be at bottommost point as in previous case but it will make an angle 60 degree with horizontal
so T1 = T2
Consider the circuit shown with key opened. 3uf capacitor is charged to p.d. of 9 V. Capacitor 6 uF is charged to p.d. of 3V. The key is now closed . What will be the potential differnce across 3 uF capacitor in steady state?
V = [3*9-6*3]/(3+6) = 9/9 = 1
here - is used for opposite poles of the capacitors are connected together
Find the magnitude of electric potential at the origrn due to following charge distribution (if q = 1 nC)
sir i am unable to solve the series :
k - 3k/2 +5k/4 - 7k/4..........
= -k/2 - 3k/8 ....... plzz help
k - 3k/2 +5k/4 - 7k/8..........
= k(1-3/2+5/4-7/8+...............)
now let S = 1-3/2+5/4-7/8+.......
and -1/2 S = -1/2+3/4-5/8+......
on substracting 2nd by 1st
we get 3/2 S = 1-1+1/2-1/4+.......
or 3/2 S = 1-(1-1/2+1/4-1/8+..........)
now solve
my previous questn from rotational topic is still uncleared . I want to know why w= 4π is used.
according to rotational mechanics angular velocity = 2πn here n is frequency and in the problem n = 120 rotation/min = 2 rotation/sec.
the time period of oscillatn of a simple pendulum is 2π (l/g)1/2. l is about 10cm & is known to 1mm accuracy. the time period of oscillatn is about 0.5s . time of 100 oscillatn with a wrist watch of 1s revolutn . what is the accuracy in determinatn of g?
ans 5%
T = 2π (l/g)1/2
so g α l/T2
so % error in g = % error in l + 2 % error in T
= 0.1*100/10 + 2* 1*100/50 = 1+4 = 5
here you shoult remember that if you count 100 osccilations then time will be counted for 100 osccilation.
from rest a body rotate 120rot/min for 20 sec then rotate for 5 sec with same rotation . find how many rotation it will make?
here while calculating ang. acc you use w= w0 + at , how w=4π , ( i want to know how to convert 120 rot/min)
120 rotation per minute = 120/60 = 2 rotation per second
A BODY STARTS WITH INITIAL VELOCITY 30m/s & A RETARDATION OF 4m/s2 . FIND THE DISTANCE TRAVELLED BY THE BODY IN 8th SECOND.
ANS 1m
by v = u - at
0 = 30 - 4t implies t = 7.5
it means velocity will become 0 in 7.5 sec.
distance covered in 7.5 sec. = 30*7.5 - 1/2*4*(7.5)2 = 225 - 225/2 = 225/2
distance covered in 7 sec. = 30*7 - 1/2*4*(7)2= 210 - 98 = 112
so distance covered in first half of eighth sec. = 112.5 - 112 = 0.5 m
distance covered in last half of eighth sec. = 0*0.5 + 1/2*4(0.5)2= 0.5
so total distance covered in eighth sec. = 0.5 + 0.5 = 1 meter
IN THE PREVIOUS QUES ( THAT WAS FROM ROTATn TOPIC , SUBMITTED 5 DAYS AGO) , I AM STILL NOT GETTING THE SOLn SUBMITTED BY YOU.
if this is the question based on number of rotations then read it carefully
A TOY CAR TRAVELS FROM A TO B AT A CONST. SPEED OF 20 km/h & IMMEDIATELY RETURNS TO A AT A CONST. SPEED v. IF THE AVG. SPEED OF THE CAR IS 24 km/h . THEN v IS = ?
total distance = d+d = 2d km.
total time taken = d/20 + d/v hours
so avg. speed = 2d/[d/20 + d/v]
or 24 = 2/[1/20 + 1/v]
or 12 = 1/[1/20 + 1/v]
or 1/20 + 1/v = 1/12
or 1/v = 1/12 - 1/20
or 1/v = 2/60
or v = 30