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A WIRE OF LENGTH L AND AREA OF CROSS SECTION A HAS A RESISTANCE R. IF THE WIRE IS STRACHED TO TWICE OF ITS LENGTH WHILE A REMAINS SAME , WHAT IS THE NEW RESISTANCE????
We know that R = ρl/A
When the wire is stretched, there is no change in volume. So if length is doubled then volume will be halved so that volume is constant.
Therefore, the new resistance R’ = 4R
That is the resistance becomes four times its original value.
1. A WIRE OF LENGTH L AND AREA OF CROSS SECTION A HAS A RESISTANCE R. IF THE WIRE IS STRACHED TO TWICE OF ITS LENGTH WHILE A REMAINS SAME , WHAT IS THE NEW RESISTANCE????
john Alexander asks
Is the energy required to break a sample of string proportional to the length of string or is the energy requirednto break the sting constant?
The energy required to break a string depends on the stress known as tensile failure stress.It is not related to the length of the string .for more knowledge ref. Mohr’s theory
Vinolia Maserumule asks:
“Find the relationship between L(the length of the shape L-shape)and S(the number of small squares in the L-shape)P(the perimeter of the L-shape)T(the total number of match sticks). Will 600 match sticks be enough to build a L-shape with sides of 100 squares?”
One mark questions with answers
(n – 1)/(1 + n) is the fraction of the momentum retained by the moving body so
(n – 1)/(1 + n) = (2 – 1)/(1 + 2) = 1/3.
e = 0.2 = v2/50, so v2 = 10 m/sec.
20 = (1/2)K (36 – 0)
W = (1/2)K (36 – 9)
Solving the above equations we get W = 15 J.
Two mark questions with answers
Because the ball is dropped from rest, hence u = 0.
Hence, v2 = u2 + 2as
= 0 + (2 x 10 x 20) = 400
So, v = 20 m/s
Kinetic energy of the ball just before hiting the ground
= (1/2)mv2 = (1/2)m(400) = 200m Joule
Because the ball loses 30% of the kinetic energy on striking the ground, hencekinetic energy retained by the ball after striking the ground = 70% of 200m J
= 140m J
The energy loss is due to the inelastic collision with the ground.
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h = H.e2n.
h = 40 x (1/2)2 x 3= 40 x (1/26) = 40/64 = 0.625 m
If force of brakes be the same then m1a1 = m2a2……….(ii)
If truck stops over a distance S1 then v12 =2a1S1 ……..(iii)
If car stops over a distance S2 then v22 = 2a2S2………(iv)
From (i) and (ii)
(1/2)m1v12 = (1/2)m2v22………..(v)
From (ii) and (v)
v12/a1 = v22/a2…………(vi)
From (iii), (iv) and (vi)
2a1S1/a1 = 2a2S2/a2.
S1 = S2
Hence distances covered S1 and S2 are equal.
If mass of water raised in one second = m kg.
Total work done in lifting water,W = mgh
Power P = W/t, but t = 1 minute = 60 sec.
4000 = mgh/60
4000 = (m x 10 x 20)/60
m = 1200 kg.
m = 2 x 3 = 6 kg
If the rod is displaced through an angle qthen the work done on it, W = mg(l/2)(1 – Cosq).
The effective length of the rod is taken to be (l/2) because in uniformdistribution of mass the centre of mass is at the geometric centre so
W = 6 x 10 x (3/2)(1 – Cos60) =45 J.
Three mark questions with answers
v2 = 2m1u1/(m1 + m2).
This is from the theory of conservation of momentum.
Momentum of the mass m2 after collision,
P2 = m2v2 = (2m1m2u1)/(m1+ m2)
Fraction of momentum transferred to m2.
= (2m1m2u1)/(m1 + m2)m1u1= 2m2/(m1 + m2)
= 2n/(1 + n) ….[because m2/m1 = n]
W = F.S,
W = FS Cos q
It remains +ve for the angle q between 0oto 90o
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Thus work is +ve if Cos q
Hence work done W = FS Cos 90o = 0.
Thus, W may be +ve, -ve or zero
= mg x 2 Joules = 2mg Joules
Kinetic energy at lowest position
= Potential energy at the highest position – the energy dissipitated againstair resistance or friction
= [mg x 2 – (10/100) x mg x2] Joule
= mg x 18/10 J
or, v = 1.9 ms-1.
W = FS
(1) In S.I system,
If F = 1 kg weight or 1 kg force and S = 1m then,
W = (1 kg wt)(1m) = 1 kg m ……………(i).
Hence, one kgm is the gravitational unit of work in S.I (M.K.S) system and isdefined as the amount of work done if 1 kg force displaces a body through 1m inthe direction of the applied force.
(2) In C.G.S system,
F = 1 gmwt and S = 1cm,
W = (1 gm wt) (1 cm) = 1 gm cm ………………(ii).
Hence, one gm cm is the gravitational unit of work and is defined as the amountof work done, if 1 gm force displaces a body through 1 cm in the direction ofthe applied force.
1 gm cm = 980 ergs.
NOTE: 1 kg m = 9.8 Joules.
One electron volt is the energy acquired by one elctron in moving it betweentwo point having a P.D of 1V.
Thus, 1eV = (1.6 x 10-19) C x1J/C = 1.6 x 10-19 Joules.
NOTE: The other practical units used are
1 Million electron volt = 1 MeV = 106 eV, 1 MeV = 106 x1.6 x 10-19 J, 1 MeV = 1.6 x10-13 Joules and
1 Billion eV = 109 eV, 1 BeV = 1.6 x 10-10joules.
Therefore, Power of water is directly proportional to the cube of velocity ofwater so let P = Kv3 (k is a constant and is equal to ‘ad’.)Taking log on both sides
log P = 3log v + log k
Differentiating on both sides
percentage change in power, DP/P x100 = 3 x 5%
r = radius of pipe = 1.2m, average speed of water v = 12 m/s
V = 240 kV = 240 x 103 volt, density ofwater p = 103 kg/m3.
Now, kinetic energy of rushing water per second i.e.
Power P = (1/2)(mass flowing per sec) x v2
= (1/2) x 3.14 x (1.2)2x 103 x (12)3watt
= 3.9 x 106 watt
current = output power/voltage
= (60% of power P)/(240 x 1000)
= [(60/100) x 3.9 x 106]/(240x 1000) = 9.75 amp.
Five mark questions with answers
(a) The velocity is given as a function of time by
v = (2Pt/m)1/2
(b) The position is given as a function of time by
s = (8P/9m)1/2t3/2.
(c) What is the shape of the graph between velocity and mass of the vehicle ifother factors remain same?
(d) What is the shape of the graph between displacement and power?
i.e., m x (dv/dt) x v =P [as F = ma = m x (dv/dt)]
After rearranging and integrating on both sides
(v2/2) = (P/m) x t + C1
Now as initially the body is at rest, i.e., v = 0 at t = 0, so C1= 0.
v = (2Pt/m)1/2 …………(1)
(b) By definition v = (ds/dt),
Using eq (1) above,
ds/dt = (2Pt/m)1/2
On integration we get
s = (2P/m)1/2 x (2/3) xt3/2 + C2.
Now, as at t = 0, s = 0, so, C2 = 0
s = (8P/9m)1/2 t3/2.
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Taking the reference level at the lowest point R, we have
hP = l – l cos q0
hQ = l – l cos q = l(1 – cos
So (a) potential energy at Q relative to R will be
PE = mghQ
PE = mgl(1 – cos q
(b) PE at P = mghP = mgl(1 – cos q0)
KE at P = 1/2 x mv2 = 0
so, total mechanical energy at P = mgl(1 – cos q0) …….(i)
Now, if KQ is the KE at Q,
then using eq. (i)
mechanical energy at Q = KQ + mgl(1 – cos
But by conservation of mechanical energy between P and Q
KQ + mgl(1 – cos q) = mgl(1 -cos q
i.e., KQ = mgl(cos q– cos q
(c) If v is the speed at point Q, from eq. (b)
1/2 x mv2 = mgl(cos
i.e., v = height=17src=”./xi%20work,%20power%20and%20energy_files/image007.gif”alt=”i2.gif (1049 bytes)” v:shapes=”_x0000_i1049″>.
(d) If ‘E’ is the energy at Ðq, then itis equal to mgl(1 – Cosq
Since the energy remains constant throughout, E = Eo.
mgl(1 – Cosq
or mv2 = 2mgl(Cosq – Cos
Therefore, tension ‘T’ at q would begiven by
T = mv2/l + mgCosq = mg Cos
or T = 3mgCosq
Q3.What do youmean by work in the language of physics? Give its absolute and gravitationalunits. Give two illustrations of zero work, negative work and positive work.
Q4.How will youfind work done by a variable force mathematically and graphically?
Q5.What do youmean by conservative and non-conservative forces? Give their importantproperties.
Q6.What do youmean by gravitational potential energy? Show that gravitational potentialenergy is independent of the path followed.
Q7.If a body iskept on the top of a rough inclined plane, find the expression for
(i) work done in bringing it down to the bottom of the plane with constantvelocity
(ii) work done in moving it up the plane with constant acceleration
(iii) work done in moving it down the plane with constant acceleration.
I am using 18 gauge nicrome wire having ohm of 17.3 , 220 volt. I want to use 10 gauge wire how to decide the length?
18 gauge corresponds to a diameter of 1.22 mm
Therefore radius r1= 1.22/2=0.61 mm=0.00061m
10 gauge means diameter is 3.25 mm
Radius r2 = 3.25/2=1.625mm=0.001625m
Sine the resistance needs to be the same, we can use the formula
As resistivity is same.
Substitute the values and you will get the answer.
A group of UK Physicists claim that they have developed graphene bubbles which can change its curvature and hence its focal length and power by applying small voltages. If developed, this can be used in mobile cameras and other such devices. This would better imitate the functioning of Human Eye too.
More details available at http://physicsworld.com/cws/article/news/47250