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Physics NCERT Exemplar Solutions Class 11th Chapter Nine

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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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Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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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Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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 NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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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Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

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)

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a)

BO+OC- (BD+DC)

= 2BO -2BD

= 2 (BO-BD)

=2 [ (x²+L²)^1/2-L]

=2L [ (1+ $\frac{x^{2}}{L^{2}}$ )^1/2-L]

= 2L [ (1+ $\frac{1}{2} \frac{x^{2}}{L^{2}} - 1$ ]= $\frac{x^{2}}{L}$

Strain = $\frac{? L}{2 L} = \frac{\frac{x^{2}}{L}}{2 L} = \frac{x^{2}}{2 L^{2}}$

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Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

A rigid bar of mass M is supported symmetrically by three wires each of length l . Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

(a) Y_copper /Y_iron

(b) $\sqrt[]{\frac{Y_{i r o n}}{Y_{c o p p e r}}}$

(c) $\frac{Y_{i r o n}^{2}}{Y_{c o p p e r}^{2}}$

(d) Y_iron/Y_copper .

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Y= $\frac{s t r e s s}{s t r a i n} = \frac{\frac{F}{A}}{\frac{? L}{L}} = \frac{F}{π ({D / 2)}^{2}} * \frac{L}{? L}$

D²= $\frac{4 F L}{π ? L Y}$

D= $\sqrt[]{\frac{4 F L}{π ? L Y}}$

$\frac{D_{c o p p e r}}{D_{i r o n}} = \sqrt[]{\frac{Y_{i r o n}}{Y_{c o p p e r}}}$

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Nine Class 11th

A spring is stretched by applying a load to its free end. The strain produced in the spring is

(a) Volumetric

(b) Shear

(c) Longitudinal and shear

(d) Longitudinal

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes. The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

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