A solid cylinder of radius r and height h
Question: A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4rh + 4r2. Solution: False Since, the total surface area of cylinder of radius, rand height, h = 2rh + 2r2When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder = 2 h and radius of the new cylinder = r Total surface area of the new cylinder = 2r(2h) + 2r2= 4rh + 2r2...
Read More →Find the product:
Question: Find the product:(x2a2) (xa) Solution: By horizontal method: $\left(x^{2}-a^{2}\right) \times(x-a)$ $=x^{2}(x-a)-a^{2}(x-a)$ $=x^{3}-a x^{2}-a^{2} x+a^{3}$ i.e $\left(x^{3}+a^{3}\right)-a x(x-a)$...
Read More →Find the product: (3m − 4n) × (2m − 3n)
Question: Find the product:(3m 4n) (2m 3n) Solution: By horizontal method: $(3 m-4 n) \times(2 m-3 n)$ $=3 m(2 m-3 n)-4 n(2 m-3 n)$ $=6 m^{2}-9 m n-8 m n+12 n^{2}$ $=6 m^{2}-17 m n+12 n^{2}$...
Read More →Two identical solid hemispheres of equal
Question: Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6r2. Solution: False Curved surface area of a hemisphere = 2 r2 Here, two identical solid hemispheres of equal radius are stuck together. So, base of both hemispheres is common. Total surface area of the combination = 2 r2+ 2 r2= 4 r2...
Read More →Find the product: (9x + 5y) × (4x + 3y)
Question: Find the product:(9x+ 5y) (4x+ 3y) Solution: By horizontal method: $(9 x+5 y) \times(4 x+3 y)$ $9 x(4 x+3 y)+5 y(4 x+3 y)$ $=36 x^{2}+27 x y+20 x y+15 y^{2}$ $=36 x^{2}+47 x y+15 y^{2}$...
Read More →If volumes of two spheres are in the ratio 64 : 27,
Question: If volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is (a) 3: 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9 Solution: (d)Let the radii of the two spheres are r1and r2, respectively. $\therefore$ Volume of the sphere of radius, $r_{1}=V_{1}=\frac{4}{3} \pi r_{1}^{3}$$\ldots$ (i) $\left[\because\right.$ volume of sphere $=\frac{4}{3} \pi$ (radius) $\left.^{3}\right]$ and volume of the sphere of radius, $r_{2}=V_{2}=\frac{4}{3} \pi r_{2}^{3}$ ...(ii) Given, ratio...
Read More →Differentiate each of the following functions from the first principal :
Question: Differentiate each of the following functions from the first principal : $\log \cos x$ Solution: We have to find the derivative of $\log \cos x$ with the first principle method, so, $f(x)=\log \cos x$ by using the first principle formula, we get, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \cos (x+h)-\log \cos x}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\cos (x+h)}{\cos x}\right)}{h}$ $f^{\prim...
Read More →Find the product: (7x + 2y) × (x + 4y)
Question: Find the product:(7x+ 2y) (x+ 4y) Solution: By horizontal method: $(7 x+2 y) \times(x+4 y)$ $=7 x(x+4 y)+2 y(x+4 y)$ $=7 x^{2}+28 x y+2 x y+8 y^{2}$ $=7 x^{2}+30 x y+8 y^{2}$...
Read More →Find the product: (3y − 8) × (5y − 1)
Question: Find the product:(3y 8) (5y 1) Solution: By horizontal method: $(3 y-8) \times(5 y-1)$ $=3 y(5 y-1)-8(5 y-1)$ $=15 y^{2}-3 y-40 y+8$ $=15 y^{2}-43 y+8$...
Read More →Find the product: (2x + 5) × (4x − 3)
Question: Find the product:(2x+ 5) (4x 3) Solution: By horizontal method: $(2 x+5) \times(4 x-3)$ $=2 x(4 x-3)+5(4 x-3)$ $=8 x^{2}-6 x+20 x-15$ $=8 x^{2}+14 x-15$...
Read More →Find the product: (4x + 9) × (x − 6)
Question: Find the product:(4x+ 9) (x 6) Solution: By horizontal method: $(4 x+9) \times(x-6)$ $=4 x(x-6)+9(x-6)$ $=4 x^{2}-24 x+9 x-54$ $=4 x^{2}-15 x-54$...
Read More →Find the product: (5x + 7) × (3x + 4)
Question: Find the product:(5x+ 7) (3x+ 4) Solution: By horizontal method: $(5 x+7) \times(3 x+4)$ $=5 x(3 x+4)+7(3 x+4)$ $=15 x^{2}+20 x+21 x+28$ $=15 x^{2}+41 x+28$...
Read More →In a right circular cone,
Question: In a right circular cone, the cross-section made by a plane parallel to the base is a (a) circle (b) frustum of a cone (c) sphere (d) hemisphere Solution: (b)We know that, if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone....
Read More →The perimeter of a triangle is
Question: The perimeter of a triangle is 6p2 4p+ 9 and two of its sides arep2 2p+ 1 and 3p2 5p+ 3. Find the third side of the triangle. Solution: Let $a, b$ and $c$ be the three sides of the triangle. $\therefore$ Perimeter of the triangle $=(a+b+c)$ Given perimeter of the triangle $=6 p^{2}-4 p+9$ One side $(a)=p^{2}-2 p+1$ Other side $(b)=3 p^{2}-5 p+3$ Perimeter $=(a+b+c)$ $\left(6 p^{2}-4 p+9\right)=\left(p^{2}-2 p+1\right)+\left(3 p^{2}-5 p+3\right)+c$ $6 p^{2}-4 p+9-p^{2}+2 p-1-3 p^{2}+5 p...
Read More →Differentiate the following functions from first principles :
Question: Differentiate the following functions from first principles : $e^{\sqrt{2 x}}$ Solution: We have to find the derivative of $\mathrm{e}^{\sqrt{2} x}$ with the first principle method, so, $f(x)=e^{\sqrt{2} x}$ by using the first principle formula, we get, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2 x}}}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{2 x}}\left(e^{\sqrt{2(x+h)}}-\sqrt{2 ...
Read More →The two adjacent sides of a rectangle are
Question: The two adjacent sides of a rectangle are 5x2 3y2andx2+ 2xy. Find the perimeter. Solution: Sides of the rectangle are $l$ and $b$. $l=5 x^{2}-3 y^{2}$ $b=x^{2}+2 x y$ Perimeter of the rectangle is $(2 l+2 b)$. Perimeter $=2\left(5 x^{2}-3 y^{2}\right)+2\left(x^{2}+2 x y\right)$ $=10 x^{2}-6 y^{2}+2 x^{2}+4 x y$ Hence, the perimeter of the rectangle is $12 x^{2}-6 y^{2}+4 x y$....
Read More →The diameters of the two circular ends
Question: The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is (a) 32.7 L (b) 33.7 L (c) 34.7 L (d) 31.7 L Solution: (a)Given, diameter of one end of the bucket $2 R=44 \Rightarrow R=22 \mathrm{~cm}$$[\because$ diameter, $r=2 \times$ radius $]$ and diameter of the other end, $2 r=24 \Rightarrow r=12 \mathrm{~cm} \quad[\because$ diameter, $r=2 \times$ radius $]$ Height of the bucket, $h=35 \mathrm{~cm}$ Since, t...
Read More →What must be subtracted from
Question: What must be subtracted from 3a2 6ab 3b2 1 to get 4a2 7ab 4b2+ 1? Solution: Let the required number bex. $\left(3 a^{2}-6 a b-3 b^{2}-1\right)-x=4 a^{2}-7 a b-4 b^{2}+1$ $\left(3 a^{2}-6 a b-3 b^{2}-1\right)-\left(4 a^{2}-7 a b-4 b^{2}+1\right)=x$ $\therefore$ Required number $=-a^{2}+a b+b^{2}-2$...
Read More →Differentiate the following functions from first principles:
Question: Differentiate the following functions from first principles: $e^{\cos x}$ Solution: We have to find the derivative of $e^{\cos x}$ with the first principle method, so, $f(x)=e^{\cos x}$ by using the first principle formula, we get, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\cos (x+h)}-e^{\cos x}}{h}$ $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\cos x}\left(e^{\cos (x+h)-\cos x}-1\right)}{h}$ [By using $\lim _{x \ri...
Read More →Subtract:
Question: Subtract:4p2+ 5q2 6r2+ 7 from 3p2 4q2 5r2 6 Solution: Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:...
Read More →During conversion of a solid from
Question: During conversion of a solid from one shape to another, the volume of the new shape will (a) increase (b) decrease (c) remain unaltered (d) be doubled Solution: (c) During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered....
Read More →Subtract:
Question: Subtract:5y4 3y3+ 2y2+y 1 from 4y4 2y3 6y2y+ 5 Solution: Arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:...
Read More →A right circular cylinder of radius
Question: A right circular cylinder of radius r cm and height h cm (where, h2r) just encloses a sphere of diameter (a) r cm (b) 2r cm (c) h cm (d) 2h cm Solution: (b)Because the sphere encloses in the cylinder, therefore the diameter of sphere is equal to diameter of cylinder which is 2r cm....
Read More →Subtract:
Question: Subtract:x3+ 3x2 5x+ 4 from 3x3x2+ 2x 4 Solution: On arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:...
Read More →Subtract: −6p + q + 3r + 8 from p − 2q − 5r − 8
Question: Subtract:6p+q+ 3r+ 8 fromp 2q 5r 8 Solution: Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:...
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