Q8. A solid sphere, of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity η. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, v2, when falling through the same fluid, the ratio ( v1v2 ) equals: (1) 1 (2) 27 9 (3) 1 (4) 9 27
What This Question Tests
This problem requires applying Stokes' law and the principle of volume conservation to determine the relationship between terminal velocities of a large sphere and smaller spheres derived from it.
Concepts Tested
Formulas Used
v_t = (2r²(ρ-σ)g) / (9η)
📚 NCERT Sections This Tests
8.2 — Stress And Strain
Physics Class 12 · Chapter 8
8.2 STRESS AND STRAIN forces are applied parallel to the cross-sectional When forces are applied on a body in such a area of the cylinder, as shown in Fig. 8.1(b), manner that the body is still in static equilibrium, there is relative displacement between the it is deformed to a small or large extent depending opposite faces of the cylinder. The restoring force upon the nature of the material of the body and per unit area developed due to the applied the magnitude of the deforming force. The tangential force is known as tangential or deformation may not be noticeable visually in shearing stress. many materials but it is there. When a body is As a result of applied tangential force, there subjected to a deforming force, a restoring force is a relative displacement ∆x between opposite is developed in the body. This restoring force is faces of the cylinder as shown in the Fig. 8.1(b). equal in magnitude but opposite in direction to The strain so produced is known as shearing the applied force. The restoring force per unit area strain and it is defined as the ratio of relative is known as stress. If F is the force applied normal displacement of the faces ∆x to the length of the to the cross–section and A is the area of cross cylinder L. section of the body, ∆x Magnitude of the stress = F/A (8.1) Shearing strain = = tan θ (8.3) L The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T–2 ]. where θ is the angular displacement of the There are three ways in which a solid may cylinder from the vertical (original position of the change its dimensions when an external force cylinder). Usually θ is very small, tan θ acts on it. These are shown in Fig. 8.1. In is nearly equal to angle θ, (if θ = 10°, for Fig.8.1(a), a cylinder is stretched by two equal example, there is only 1% difference between θ forces applied normal to its cross-sectional area. and tan θ). The restoring force per unit area in this case is It can also be visualised, when a book is called tensile stress. If the cylinder is pressed with the hand and pushed horizontally, compressed under the action of applied forces, as shown in Fig. 8.2 (c). the restoring force per unit area is known as Thus, shearing strain = tan θ ≈ θ (8.4) compressive stress. Tensile or compressive In Fig. 8.1 (d), a solid sphere placed in the fluid stress can also be termed as longitudinal stress. under high pressure is compressed uniformly on In both the cases, there is a change in the all sides. The force applied by the fluid acts in length of the cylinder. The change in the length perpendicular direction at each point of the ∆L to the original length L of the body (cylinder surface and the body is said to be under in this case) is known as longitudinal strain. hydraulic compression. This leads to decrease (a) (b) (c) (d) Fig. 8.1 (a) A cylindrical body under tensile stress elongates by ∆L (b) Shearing stress on a cylinder deforming it by an angle θ(c) A body subjected to shearing stress (d) A solid body under a stress normal to the surface at every point (hydraulic stress). The volumetric strain is ∆V/V, but there is no change in shape. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 169 in its volume without any change of its compression and shear stress may also be geometrical shape. obtained. The stress-strain curves vary from The body develops internal restoring forces material to material. These curves help us to that are equal and opposite to the forces applied understand how a given material deforms with by the fluid (the body restores its original shape increasing loads. From the graph, we can see and size when taken out from the fluid). The that in the region between O to A, the curve is internal restoring force per unit area in this case linear. In this region, Hooke’s law is obeyed. is known as hydraulic stress and in magnitude The body regains its original dimensions when is equal to the hydraulic pressure (applied force the applied force is removed. In this region, the per unit area). solid behaves as an elastic body. The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (∆V) to the original volume (V). ∆V Volume strain = (8.5) V Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula.
9.31 — Figure 9.30 Shows An Equiconvex Lens (Of Refractive Index 1.50) In
Physics Class 12 · Chapter 9
9.31 Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? FIGURE 9.30 253 Reprint 2025-26 Physics Notes Reprint 2025-26 Wave Optics Chapter Ten WAVE OPTICS 10.1 INTRODUCTION In 1637 Descartes gave the corpuscular model of light and derived Snell’s law. It explained the laws of reflection and refraction of light at an interface. The corpuscular model predicted that if the ray of light (on refraction) bends towards the normal then the speed of light would be greater in the second medium. This corpuscular model of light was further developed by Isaac Newton in his famous book entitled OPTICKS and because of the tremendous popularity of this book, the corpuscular model is very often attributed to Newton. In 1678, the Dutch physicist Christiaan Huygens put forward the wave theory of light – it is this wave model of light that we will discuss in this chapter. As we will see, the wave model could satisfactorily explain the phenomena of reflection and refraction; however, it predicted that on refraction if the wave bends towards the normal then the speed of light would be less in the second medium. This is in contradiction to the prediction made by using the corpuscular model of light. It was much later confirmed by experiments where it was shown that the speed of light in water is less than the speed in air confirming the prediction of the wave model; Foucault carried out this experiment in 1850. The wave theory was not readily accepted primarily because of Newton’s authority and also because light could travel through vacuum255 Reprint 2025-26 Physics and it was felt that a wave would always require a medium to propagate from one point to the other. However, when Thomas Young performed his famous interference experiment in 1801, it was firmly established that light is indeed a wave phenomenon. The wavelength of visible light was measured and found to be extremely small; for example, the wavelength of yellow light is about 0.6 mm. Because of the smallness of the wavelength of visible light (in comparison to the dimensions of typical mirrors and lenses), light can be assumed to approximately travel in straight lines. This is the field of geometrical optics, which we had discussed in the previous chapter. Indeed, the branch of optics in which one completely neglects the finiteness of the wavelength is called geometrical optics and a ray is defined as the path of energy propagation in the limit of wavelength tending to zero. After the interference experiment of Young in 1801, for the next 40 years or so, many experiments were carried out involving the interference and diffraction of lightwaves; these experiments could only be satisfactorily explained by assuming a wave model of light. Thus, around the middle of the nineteenth century, the wave theory seemed to be very well established. The only major difficulty was that since it was thought that a wave required a medium for its propagation, how could light waves propagate through vacuum. This was explained when Maxwell put forward his famous electromagnetic theory of light. Maxwell had developed a set of equations describing the laws of electricity and magnetism and using these equations he derived what is known as the wave equation from which he predicted the existence of electromagnetic waves*. From the wave equation, Maxwell could calculate the speed of electromagnetic waves in free space and he found that the theoretical value was very close to the measured value of speed of light. From this, he propounded that light must be an electromagnetic wave. Thus, according to Maxwell, light waves are associated with changing electric and magnetic fields; changing electric field produces a time and space varying magnetic field and a changing magnetic field produces a time and space varying electric field. The changing electric and magnetic fields result in the propagation of electromagnetic waves (or light waves) even in vacuum. In this chapter we will first discuss the original formulation of the Huygens principle and derive the laws of reflection and refraction. In Sections 10.4 and 10.5, we will discuss the phenomenon of interference which is based on the principle of superposition. In Section 10.6 we will discuss the phenomenon of diffraction which is based on Huygens- Fresnel principle. Finally in Section 10.7 we will discuss the phenomenon of polarisation which is based on the fact that the light waves are transverse electromagnetic waves. * Maxwell had predicted the existence of electromagnetic waves around 1855; it was much later (around 1890) that Heinrich Hertz produced radiowaves in the laboratory. J.C. Bose and G. Marconi made practical applications of the Hertzian 256 waves Reprint 2025-26 Wave Optics
9.27 — (A) M = ( Fo/Fe) = 28
Physics Class 12 · Chapter 9
9.27 (a) m = ( fO/fe) = 28 f O f O (b) m = 1 + = 33.6 f e 25 349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2 0.28 10 – 3 10.4 m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers
📋 Question Details
- Chapter
- Properties of Matter
- Topic
- Viscosity
- Year
- 2019
- Shift
- 12 Apr Shift 2
- Q Number
- Q8
- Type
- MCQ
- NCERT Ref
- Class 11 Physics Ch 10: Mechanical Properties of Fluids
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