Prove that
Question: Prove that $\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=1$ Solution: To Prove: $\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=1$ Taking LHS, $=\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}$ Multiply and divide by tan 54 tan 18 $=\frac{\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}}{\tan 54^{\circ} \tan 18^{\circ}} \times \tan 54^{\circ} \tan 18^{\circ}$ $=\frac{\left(\tan 6^{\circ} \tan 54^{\circ} \tan 66^{\circ}\righ...
Read More →(a) A steel wire of mass μ per unit length
Question: (a) A steel wire of mass per unit length with a circular cross-section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg/m3. (b) If the yield strength of steel is 2.5 108N/m2, what is the maximum weight that can be ...
Read More →Prove that
Question: Prove that $\sin ^{2} 72^{\circ}-\cos ^{2} 30^{\circ}=\frac{(\sqrt{5}-1)}{8}$ Solution: To Prove: $\sin ^{2} 72^{\circ}-\cos ^{2} 30^{\circ}=\frac{\sqrt{5}-1}{8}$ Taking LHS, $=\sin ^{2} 72^{\circ}-\cos ^{2} 30^{\circ}$ $=\sin ^{2}\left(90^{\circ}-18^{\circ}\right)-\cos ^{2} 30^{\circ}$ $=\cos ^{2} 18^{\circ}-\cos ^{2} 30^{\circ} \ldots$ (i) Here, we dont know the value of cos 18. So, we have to find the value of cos 18 Let $x=18^{\circ}$ so, $5 x=90^{\circ}$ Now, we can write $2 x+3 x...
Read More →Two identical solid balls,
Question: Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the flood and why? Solution: Ivory ball will rise to a greater height after striking the flood because the ivory ball is more elastic than clay ball and the amount of energy and momentum transferred is more on the ivory ball....
Read More →Find the point on the curve
Question: Find the point on the curve $y=x^{2}$ where the slope of the tangent is equal to the $x$-coordinate of the point. Solution: Let the required point be (x1,y1).Given: $y=x^{2}$ Point $\left(x_{1}, y_{1}\right)$ lies on a curve. $\therefore y_{1}=x_{1}^{2} \quad \ldots(1)$ Now, $y=x^{2} \Rightarrow \frac{d y}{d x}=2 x$ Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=2 x_{1}$ Slope of the tangent $=x$ coordinate of the point $[$ G...
Read More →Find the point on the curve
Question: Find the point on the curve $y=x^{2}$ where the slope of the tangent is equal to the $x$-coordinate of the point. Solution: Let the required point be (x1,y1).Given: $y=x^{2}$ Point $\left(x_{1}, y_{1}\right)$ lies on a curve. $\therefore y_{1}=x_{1}^{2} \quad \ldots(1)$ Now, $y=x^{2} \Rightarrow \frac{d y}{d x}=2 x$ Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=2 x_{1}$ Slope of the tangent $=x$ coordinate of the point $[$ G...
Read More →A truck is pulling a car out of a ditch
Question: A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? Solution: Length of steel cable, l = 9.1 m Radius, r = 5 mm Tension on the cable = 800 N Youngs modulus = Y = 2 1011N/m2 Y = (F/A)/( ∆L/L) ∆L = 0.5 10-3m...
Read More →To what depth must a rubber
Question: To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1%. Solution: Bulk modulus of rubber = 9.8 108N/m2 Density of sea water = 103kg/m3 Percentage decrease in volume = 0.1% Change in pressure, p = hg B = ∆P/(∆V/V) ∆P = hg h = 100 m...
Read More →A steel rod of length 1 m and area of cross-section
Question: A steel rod of length 1 m and area of cross-section 1 cm2is heated from 0oC to 200oC without being allowed to extend or bend. What is the tension produced in the rod? Solution: Youngs modulus of steel, Y = 2.0 1011N/m2 Coefficient of thermal expansion, = 10-5oC-1 Length, L = 1 m Area of cross section A = 1 cm2= 110-4m2 Rise in temperature = 200oC Tension produced in rod by the compression is Y = (T/A)/(∆L/L) T = 4 104N...
Read More →What is the Bulk modulus
Question: What is the Bulk modulus for a perfect rigid body? Solution: Bulk modulus is given as: Bulk modulus = Stress/ Volume strain = pV/∆V For a perfectly rigid body, bulk modulus is B = pV/0 =...
Read More →What is the Young’s modulus
Question: What is the Youngs modulus for a perfect rigid body? Solution: According to Hookes law, Youngs modulus = Stress/ Longitudinal strain When the given body is a perfect rigid body, then the Youngs modulus is given as: $Y=\frac{F}{A} \times \frac{l}{0}=\infty$...
Read More →Find the points on the curve
Question: Find the points on the curve $x y+4=0$ at which the tangents are inclined at an angle of $45^{\circ}$ with the $x$-axis. Solution: Let the required point be (x1,y1).Slope of the tangent at this point = tan 45= 1Given: $x y+4=0 \ldots(1)$ Since the point satisfies the above equation, $x_{1} y_{1}+4=0 \quad \ldots(2)$ On differentiating equation (2) both sides with respect to $x$, we get $x \frac{d y}{d x}+y=0$ $\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$ Slope of the tangent at $\left(x_{...
Read More →Identical springs of steel and copper
Question: Identical springs of steel and copper are equally stretched. On which, more work will have to be to done? Solution: When the identical springs of steel and copper are equally stretched, force is given as: $W \propto \Delta l$ Because the springs are same, their length and area will also be the same $\Delta l \propto \frac{l}{Y}$ From both the equations, we can say that $\frac{W_{\text {steel }}}{W_{\text {copper }}}=\frac{Y_{\text {copper }}}{Y_{\text {steel }}} \Rightarrow W_{\text {s...
Read More →Find the points on the curve
Question: Find the points on the curve $x y+4=0$ at which the tangents are inclined at an angle of $45^{\circ}$ with the $x$-axis. Solution: Let the required point be (x1,y1).Slope of the tangent at this point = tan 45= 1Given: $x y+4=0 \ldots(1)$ Since the point satisfies the above equation, $x_{1} y_{1}+4=0 \quad \ldots(2)$ On differentiating equation (2) both sides with respect to $x$, we get $x \frac{d y}{d x}+y=0$ $\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$ Slope of the tangent at $\left(x_{...
Read More →Is stress a vector quantity?
Question: Is stress a vector quantity? Solution: Stress is a neither a scalar quantity nor a vector quantity. It is a tensor quantity. This is because stress is equal to the magnitude of the internal reaction force divided by the area of cross-section....
Read More →The Young’s modulus for steel is more
Question: The Youngs modulus for steel is more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress? Solution: Youngs modulus is given as: Young's modulus $=\frac{\text { Tensile stress }}{\text { Longitudinal strain }}$ When the longitudinal strain is the same for both, steel will have a greater tensile stress. $\frac{Y_{\text {steel }}}{Y_{\text {rubber }}}=\frac{\text { Stress }_{\text {steel }}}{\text { Stress_rubber }}$ $\frac{Y_{\text {steel }}...
Read More →Prove that
Question: Prove that $\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}=\frac{(\sqrt{5}-1)}{8}$ Solution: To Prove: $\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}=\frac{\sqrt{5}-1}{8}$ Taking LHS, $=\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}$ We know that, $\sin ^{2} A-\sin ^{2} B=\sin (A+B) \sin (A-B)$ $=\sin \left(24^{\circ}+6^{\circ}\right) \sin \left(24^{\circ}-6^{\circ}\right)$ $=\sin 30^{\circ} \sin 18^{\circ} \ldots(\mathrm{i})$ Now, we will find the value of sin 18 Let $x=18^{\circ}$ so, $5 x=90^{\circ}...
Read More →A copper and a steel wire of the same diameter
Question: 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 1 cm. The two wires will have (a) the same stress (b) different stress (c) the same strain (d) different strain Solution: The correct answers are (a) the same stress (d) different strain...
Read More →For an ideal liquid
Question: 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 Solution: The correct answers are (a) the bulk modulus is infinite (d) the shear modulus is zero...
Read More →Find the points on the curve
Question: Find the points on the curve $y^{2}=2 x^{3}$ at which the slope of the tangent is 3 . Solution: Let $\left(x_{1}, y_{1}\right)$ be the required point. Given: $y^{2}=2 x^{3}$ Since $\left(x_{1} y_{1}\right)$ lies on a curve, $y_{1}^{2}=2 x_{1}{ }^{3}$ .....(i) $\Rightarrow 2 y \frac{d y}{d x}=6 x^{2}$ $\Rightarrow \frac{d y}{d x}=\frac{6 x^{2}}{2 y}=\frac{3 x^{2}}{y}$ Slope of the tangent at $(x, y)=\frac{3 x_{1}{ }^{2}}{y_{1}}$ Slope of the tangent $=3$ [Given] $\therefore \frac{3 x_{1...
Read More →A wire is suspended from the ceiling and
Question: 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 Solution: The correct answers are (a) tensile stress at any cross-section A of ...
Read More →Find the points on the curve
Question: Find the points on the curve $y=x^{3}-2 x^{2}-2 x$ at which the tangent lines are parallel to the line $y=2 x-3$. Solution: Let $\left(x_{1}, y_{1}\right)$ be the required point. Given: $y=2 x-3$ $\therefore$ Slope of the line $=\frac{d y}{d x}=2$' $y=x^{3}-2 x^{2}-2 x$ Since $\left(x_{1} y_{1}\right)$ lies on curve, $y_{1}=x_{1}{ }^{3}-2 x_{1}{ }^{2}-2 x_{1} \ldots(1)$ $\Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 x_{1}^{2}-4 x_{1}-2$ It is given that the tang...
Read More →Prove that
Question: Prove that $3 \sin 40^{\circ}-\sin ^{3} 40^{\circ}=\frac{\sqrt{3}}{2}$ Solution: To prove: $3 \sin 40^{\circ}-\sin ^{3} 40^{\circ}=\frac{\sqrt{3}}{2}$ Taking LHS, $=3 \sin 40^{\circ}-\sin ^{3} 40^{\circ} \ldots$ (i) We know that $\sin 3 x=3 \sin x-\sin ^{3} x$ Here, x = 40 So, eq. (i) becomes $=\sin 3\left(40^{\circ}\right)$ $=\sin 120^{\circ}$ $=\sin \left(180^{\circ}-60^{\circ}\right)$ $=\sin 60^{\circ}\left[\because \sin \left(180^{\circ}-\theta\right)=\sin \theta\right]$ $=\frac{\s...
Read More →Find a point on the curve
Question: Find a point on the curve $y=x^{3}-3 x$ where the tangent is parallel to the chord joining $(1,-2)$ and $(2,2)$. Solution: Let (x1,y1) be the required point. Slope of the chord $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2+2}{2-1}=4$ $y=x^{3}-3 x$ $\Rightarrow \frac{d y}{d x}=3 x^{2}-3 \ldots(1)$ Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 \mathrm{x}_{1}^{2}-3$ It is given that the tangent and the chord are parallel. $\therefore$ Slope of the tangent $...
Read More →Find a point on the curve
Question: Find a point on the curve $y=x^{3}-3 x$ where the tangent is parallel to the chord joining $(1,-2)$ and $(2,2)$. Solution: Let (x1,y1) be the required point. Slope of the chord $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2+2}{2-1}=4$ $y=x^{3}-3 x$ $\Rightarrow \frac{d y}{d x}=3 x^{2}-3 \ldots(1)$ Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 \mathrm{x}_{1}^{2}-3$ It is given that the tangent and the chord are parallel. $\therefore$ Slope of the tangent $...
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