2018-10-01_Physics_For_You

where k

d

m

=

ρ

is a constant of the wire. Taking

logarithm of both sides of (i) we have

log R = logk + 2 logl

Dierentiating

δδR δ

R

l

l

l

l

=+ 0 =

22

Given

δl

l

= 02 .%.

erefore,

δR

R

=× 20 .% 2 = 0.4 %.

us, the resistance of the increases by 0.2 %.

25. (d) : e time period of simple pendulum is given
    by

T l

g

= 2 π

or T lor

g

g l

T

2 22

2

==^44 ππ

As 4 and p are constants, maximum permissible

error in g is given by

∆g ∆∆

g

l

l

T

T

=+^2

Here 'L = 0.1 cm, L = 1 m = 100 cm,

'T = 0.1 s T = 50 s

∴=+  =+ 

∆g

g

01

100

2 01

50

01

100

01

25

....

or ..

∆g

g

×= +







100 01 ×

100

01

25

100

= 0.1 + 0.4 = 0.5%

26. (c) : Percentage error in T,
    ∆∆T ∆ ∆
    T

r

r

g

g

h

h

%=+%%%+

=×^001 +×+×

125

100 001

980

100 001

145

. 100

.

.

.

.

.

= 0.8 + 0.1 + 0.7 = 1.6

27. (b) : Young's modulus =

stress

strain

or Y Fl

dl

== ×× ×

××−−××

4 4198 2

ππ^23 ∆ 04 10 2308 10

(.)

(. )(.)

= 2 × 10^11 N m–2.

Now,

∆∆∆∆

∆

Y

Y

d

d

l

l

=+

2()

or ∆Y=

×

+







××

(^2001) −
04

005

08

21011 2

.

.

.

.

Nm

= 0.2 × 10^11 N m–2

Hence the Young's modulus obtained from the

reading is (2.0 r 0.2) × 10^11 N m–2.

28. (c) : Least count of screw gauge =

Pitch

n

=

05

50

001

.

=. mm

? Diameter of ball

= (2 × 0.5 mm) + (25 – 5) × 0.01 mm

= 1 + (0.01 × 20) = 1.2 mm.

29. (c) :

When a metre bridge is balanced, then

R

x

R

x

12

100

=

()−

...(i)

From gure,

R 1 = R, R 2 = 90 :, x = 40 cm

en, R

40

90

100 40

90

60

=

−

=

()

Ÿ R = 60 :

Now, for 'R taking natural log on both sides of eqn. (i),

lnR 1 = lnR 2 + lnx –ln(100 – x)

or lnR = lnx – ln(100 – x) + ln(90)

On dierentiating,

∆∆R ∆

R

x

x

x

x

=−

−

−

()

()

100

100

Ÿ

∆∆R ∆

R

x

x

x

x

=+

()100−

Ÿ ∆ΩR=+







×=×=

01

40

01

60

60

05

120

60 025

...

.

? Required value of R = (60 ± 0.25) :

30. (d) : (a) Intensity of the fundamental is more than
    that of the overtones. erefore the 1st resonance
    was having more intensity.
    (b) e prongs should not be in the
    horizontal position but vertical over
    the resonance tube.
    (c) e antinodes are formed always
    a little above the open end of the
    tube.
    is is called end correction.
    is eect will be there for overtones also.
    ? Length of the air column is less than l/4.
    