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Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density ( 1/P dp/dt ) is constant. The velocity v of any point on the surface of the expanding sphere is proportional to
12
Oct
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density ( 1/P dp/dt ) is constant. The velocity v of any point on the surface of the expanding sphere is [...]
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D = 1.25 × 10^−2 m and rise of water in capillary ,
h = 1.45 × 10^−2 m. Taking g = 9.80 m/s^2 and using the relation T = (rgh/2) × 10^3 N/m ,
The following observations were taken for dtermining the surface tension of water by capillary tube method: diameter of capillary ,
what is the possible error in measurement of surface tension T? ,
A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a high h, where the radius of its cross section is b. If the surface tension of water is S, its density if ρ, and its contact angle with glass is θ, the value of h will be (g is the acceleration due to gravity)
12
Oct
A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a high h, where the radius of its cross section is b. If the surface tension of [...]
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A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sections of different radii. When dipped in water vertically ,
and its contact angle with glass is θ ,
its density if ρ ,
the value of h will be (g is the acceleration due to gravity) ,
water rises in it to a high h ,
where the radius of its cross section is b. If the surface tension of water is S ,
3. For each given solid, identify the top view, front view and side view.
12
Oct
3. For each given solid, identify the top view, front view and side view. 3. For each given solid front view and side view. identify the top view October 12, 2020 Category: Chapter 10 - Visualising Solid Shapes , Maths , NCERT Class 8 ,
8. Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (iv) 9.7 x 9.8
12
Oct
8. Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (iv) 9.7 x 9.8 8. Using (x + a) (x + b) = x^2 + (a + b) x + ab find (iv) 9.7 x 9.8 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities [...]
8. Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (iii) 103 x 98
12
Oct
8. Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (iii) 103 x 98 8. Using (x + a) (x + b) = x^2 + (a + b) x + ab find (iii) 103 x 98 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities [...]
8. Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (ii) 5.1 x 5.2
12
Oct
8. Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (ii) 5.1 x 5.2 8. Using (x + a) (x + b) = x^2 + (a + b) x + ab find (ii) 5.1 x 5.2 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities [...]
7. Using a^2 – b^2 = (a + b) (a – b), find (iv) 12.1^2– 7.9^2
12
Oct
7. Using a^2 – b^2 = (a + b) (a – b), find (iv) 12.1^2– 7.9^2 7. Using a^2 – b^2 = (a + b) (a – b) find (iv) 12.1^2– 7.9^2 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities , Maths , NCERT Class 8 ,
7. Using a^2 – b^2 = (a + b) (a – b), find (iii) 153^2– 147^2
12
Oct
7. Using a^2 – b^2 = (a + b) (a – b), find (iii) 153^2– 147^2 7. Using a^2 – b^2 = (a + b) (a – b) find (iii) 153^2– 147^2 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities , Maths , NCERT Class 8 ,
7. Using a^2 – b^2 = (a + b) (a – b), find (ii) (1.02)2– (0.98)2
12
Oct
7. Using a^2 – b^2 = (a + b) (a – b), find (ii) (1.02)2– (0.98)2 7. Using a^2 – b^2 = (a + b) (a – b) find (ii) (1.02)2– (0.98)2 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities , Maths , NCERT Class 8 ,
7. Using a^2 – b^2 = (a + b) (a – b), find (i) 51^2– 49^2
12
Oct
7. Using a^2 – b^2 = (a + b) (a – b), find (i) 51^2– 49^2 7. Using a^2 – b^2 = (a + b) (a – b) find (i) 51^2– 49^2 October 12, 2020 Category: Chapter 9 - Algebraic Expressions and Identities , Maths , NCERT Class 8 ,