Answer:
true ............................................
Let original length be L. The new length is therefore 4L.
Let original cross sectional surface area of the wire be equal to πr^2.
This means original volume was L x πr^2 = Lπr^2
The volume is the same but the length is different so 4L x new surface area must be equal to Lπr^2. Let new surface area be equal to Y.
4L x Y = Lπr^2
=> Y = (πr^2 )/ 4
Using the resistivity formula,
R = pL/A. p which is resistivity is a constant so it stays the same
But this time, instead of L we have 4L and instead of πr^2 we have (πr^2)/4.
so the new resistance
= (4Lp)/ {(πr^2)/4}
= 16 (pL)/(πr^2)
= 16 (pL)/A. because πr^2 is A
since pL/A is equal to R from the formula, this is equal to
16 R.
R was 10 ohms
therefore new resistance is 16 x 10 = 160 ohms
Answer:
a) The wavelength of light is 6×10^-7 m.
b) The size of the central peak is 6×10^-3 m.
Explanation:
let:
α be the distance between the 1st and 2nd maxima
d = 0.4 mm be the grating slit space
X be the distance to the screen
a) α = λ×X/d
λ = α×d/X
= (0.3×10^-2)×(0.4×10^-3)/(2)
= 6×10^-7 m
Therefore, the wavelength of light is 6×10^-7 m.
b) the size of the central peak is given by:
D = 2×λ×X/d
= 2×(6×10^-7)×(2)/(0.4×10^-3)
= 6×10^-3 m
Therefore, the size of the central peak is 6×10^-3 m.
