Recent content by DriggyBoy

You guys mean to say, Placing the sensitive instrument inside a conductor ?

I was shuffling through the previous years question papers & I got this question ! '' A sensitive instrument is influenced by the strong electric field. Write a possible way to prevent this effect. Why is the electric field normal to the surface? '' I am not sure what the answer it. But this...

The question is, Is the K.E. energy conserved during the collision ? Given Data : Collision is elastic, ball of same mass. P.S. It is asked if the K.E. is conserved DURING the impact; not before or after it. After the collision & before it, I know the K.E. is conserved.

R_1 is 2 ohm nd R_2 is 4 ohm

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In the picture that i attached i need some help ******** In the book it says ''' The negative terminal of the battery E_1 is connected to b. So at the lower end of the resistance R_2 the pd is 6V. "' What is the significance of this statement?

No actually, current flows from higher potential to lower potential...so when the current reaches the resistance it is at a higher potential & whn it leaves, potential decreases... Now in the book the dirn of current is considered oppo...that is from lower to higher potential...in this case...

Does the potential decrease when it charges(electricity) pass through a resistance? Newbie here...please help :)

thnx :) now can u help me solve the prolem ! :) nd yes they are suspended from a common point by a silk thread of length 'l'

Two small eqlly chrged spheres ech of mass m, are suspnded frm same point by silk threads of length l. The distnce betwen sphres x<<l. Fnd dq/dt wth whch the charge leaks off each sphere.. If their approach velocity varies as v=a/root ovr x where a is a constnt. ***** My ques is; how can the...

Sure ^_^ For the Range, the Time of Flight and the Max height :)

A shell is fired vertically upwards with a velocity of v1 from a ship moving with a velocity v2, a person on the shore finds the motion of the shell parabola. Show that the horizontal range is 2v1v2/g

Btw i think i proceeded a bit further...Now magnification is very small... Now size of image/size of object = magnification(very small) or, size of image = magnification * size of object or, k(R^n) = magn * size of objct or, magn = { k(R^n) } / Size of objct This magn...

Cannot proceed... 1/v + 1/u = R/2 Now can we omit 1/u cause its too small? Then eqn bcums v=2/R...from this it is clear that image is at focus Now what ?

Ok thnx...but what if the object is at infinite position?? ...btw answer given in the book is 1 n=1