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Physics physics ncert exemplar solution class 12th chapter one

Five charges,q each are placed at the comers of a regular pentagon of side a.

i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the comers (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by – q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its comers?

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- (a)

(i) The electric field at the center of pentagon is zero because the distance from the center is same.

(ii) The field through one charge is Kq/r²

(iii) When one charge is positive and other is negative then net force towards negative charge. So net force is Kq/r²+ Kq/r²= 2Kq/r²

(b) It doesn't depend upon the number of sides increasing the net electric field is zero.

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Physics physics ncert exemplar solution class 12th chapter one

Five charges,q each are placed at the comers of a regular pentagon of side a.

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A copper and a steel wire of the same diameter are connected end to end. A deforming force F is applied to this composite wire which causes a total elongation of 1cm. The two wires will have

(a) The same stress.

(b) Different stress.

(c) The same strain.

(d) Different strain.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) for steel wire Y_steel= stress/strain= $\frac{f / A}{s t a r i n}$

When F and A are same for both the wires . hence stress will be same for both the wire

(Strain)_steel= stress/Y_steel and strain_copper=stress/Y_copper

Y_{steel
$\neq$}Y_copper

hence they both have different starin

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

For an ideal liquid

(a) The bulk modulus is infinite.

(b) The bulk modulus is zero.

(c) The shear modulus is infinite.

(d) The shear modulus is zero.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) An ideal liquid is not compressible

Hence $?$ V=0

Bulk modulus B= strss /volumetric strain= $\frac{\frac{F}{A}}{\frac{? V}{V}} = \infty$

Compressibility K= 1/B=1/ $\infty = 0$

As there is no tangential force exists. So shear strain =0

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. Y_Al=70 $* 10^{9} N m^{- 2}$ and Y_steel= 200 $* 10^{9} N m^{- 2}$

(a) Mass m should be suspended close to wire A to have equal stresses in both the wires

(b) Mass m should be suspended close to B to have equal stresses in both the wires

(c) Mass m should be suspended at the middle of the wires to have equal stresses in both the wires

(d) Mass m should be suspended close to wire A to have equal strain in both wires

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) Let mass m is placed at x from the end B respectively.

T_A and T_B be the tensions in wire A and wire B respectively.

For the rotational equilibrium of the system,

\sum τ = 0

T_Bx-T_A(l-x)=0

$\frac{T_{B}}{T_{A}}$ = $\frac{l - x}{x}$

Stress in wire A = S_A= $\frac{T_{A}}{a_{A}}$

Stress in wire B = S_B= $\frac{T_{B}}{a_{B}}$ where a are the area of wire

We know that a_B=2a_A

Now for equal stress

S_A=S_B

$\frac{T_{A}}{a_{A}} = \frac{T_{B}}{a_{B}}$

$\frac{T_{B}}{T_{A}} = 2$

$\frac{l - x}{x} = 2$

So x =l/3 and l-x= 2l/3

Hence mass m should placed to B.

For equal strain

Strain_A= Strain_B

$\frac{Y_{A}}{S_{A}} = \frac{Y_{B}}{S_{B}}$

$\frac{Y_{s t e e l}}{Y_{A l}} = \frac{T_{A}}{T_{B}} * \frac{a_{B}}{a_{A}} = \frac{x}{l - x} (\frac{2 a_{A}}{a_{A}})$

$\frac{200 * 10^{9}}{70 * 10^{9}} = \frac{x}{l - x}$

After solving we get x= x= 10l/17

l-x=l=10l/17=7l/17

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.

(a) Tensile stress at any cross section A of the wire is F/A

(b) Tensile stress at any cross section is zero

(c) Tensile stress at any cross section A of the wire is 2F/A

(d) Tension at any cross section A of the wire is F

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) Forces at cross section is F.

Now applying formula . stress = tension/area=F/A

Tension = applied force =F

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Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

The stress-strain graphs for two materials are shown in Fig.9.3 (assume same scale).

(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle.

(b) Material (i) and (ii) have the same elasticity and the same brittleness.

(c) Material (ii) is elastic over a larger region of strain as compared to (i).

(d) Material (ii) is more brittle than material (i).

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c), (d) The ultimate tensile strength for material ii is greater hence material ii is elastic over larger region as compared to material (i) for material (ii) fracture point is nearer, hence it is more brittle.

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods.

(a) Both the rods will elongate but there shall be no perceptible change in shape.

(b) The steel rod will elongate and change shape but the rubber rod will only elongate.

(c) The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.

(d) The steel rod will elongate, without any perceptible change in shape, but the r

...more

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) a mass M is attached at the centre. As the mass is attached to both the rods, both rod will be elongated, but due to different elastic properties of material rubber changes shape also.

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (Fig. 9.2). It can be done in one of the following three ways The tension in the strings will be

(a) The same in all cases

(b) Least in (a)

(c) Least in (b)

(d) Least in (c)

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

2Tsin $θ$ =mg

Total horizontal forces = Tcos $θ - T c o s θ = 0$

T=mg/2sin $θ$

As mg is constant T $\propto \frac{1}{s i n θ}$

T_max= mg/sin $θ$ _min

Sin $θ$ _min=0, $θ$ _min= 0

T_min=mg/2sin $θ$ _max

_{$s i n θ$
max}= 1, $θ$ =90⁰

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a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Nine

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (Fig. 9.1). A mass m is suspended from the midpoint of the wire. Strain in the wire is

(a)

(b) x/L

(c) x²/2L

(d)