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Solve the following equations: (e) 0 = 16 + 4(m – 6)
08
Oct
Solve the following equations: (e) 0 = 16 + 4(m – 6) Solve the following equations: (e) 0 = 16 + 4(m - 6) October 8, 2020 Category: Chapter 4 - Simple Equations , Maths , NCERT Class 7 ,
Solve the following equations: (d) 4 + 5(p-1) = 34
08
Oct
Solve the following equations: (d) 4 + 5(p-1) = 34 Solve the following equations: (d) 4 + 5(p-1) = 34 October 8, 2020 Category: Chapter 4 - Simple Equations , Maths , NCERT Class 7 ,
A convex spherical refracting surface with radius R separates a medium having refractive index 5/2 from air. As an object is moved towards the surface from far away from the surface along
08
Oct
A convex spherical refracting surface with radius R separates a medium having refractive index 5/2 from air. As an object is moved towards the surface from far away from the surface along A convex spherical refracting surface with radius R separates a medium having refractive index 5/2 from air. As an object is moved towards [...]
A concave refractive surface of a medium having refractive index mu produces a real image of an object (located outside the medium) irrespective of its location. Choose the correct
08
Oct
A concave refractive surface of a medium having refractive index mu produces a real image of an object (located outside the medium) irrespective of its location. Choose the correct A concave refractive surface of a medium having refractive index mu produces a real image of an object (located outside the medium) irrespective of its location. [...]
The given lens is broken into four parts rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is
08
Oct
The given lens is broken into four parts rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is The given lens is broken into four parts rearranged as shown. If the initial focal length is f then after rearrangement the equivalent focal length is October 8, 2020 [...]
Solve the following equations: (c)16 = 4 + 3(t + 2)
08
Oct
Solve the following equations: (c)16 = 4 + 3(t + 2) Solve the following equations: (c)16 = 4 + 3(t + 2) October 8, 2020 Category: Chapter 4 - Simple Equations , Maths , NCERT Class 7 ,
Solve the following equations: (b) -4 = 5(p – 2)
08
Oct
Solve the following equations: (b) -4 = 5(p – 2) Solve the following equations: (b) -4 = 5(p - 2) October 8, 2020 Category: Chapter 4 - Simple Equations , Maths , NCERT Class 7 ,
In, figure, a point object O is placed in air. A spherical boundary separates two media. AB is the principal axis. The refractive index above AB is 1,=.6 and below AB is 2.0. The separation
08
Oct
In, figure, a point object O is placed in air. A spherical boundary separates two media. AB is the principal axis. The refractive index above AB is 1,=.6 and below AB is 2.0. The separation =.6 and below AB is 2.0. The separation a point object O is placed in air. A spherical boundary separates [...]
Solve the following equations: (a) 4 = 5(p – 2)
08
Oct
Solve the following equations: (a) 4 = 5(p – 2) Solve the following equations: (a) 4 = 5(p - 2) October 8, 2020 Category: Chapter 4 - Simple Equations , Maths , NCERT Class 7 ,
The distance between an object and the screen is 100 cm. A lens produces an image on the screen when the lens is placed at either of the positions 40 cm apart. The power of the
08
Oct
The distance between an object and the screen is 100 cm. A lens produces an image on the screen when the lens is placed at either of the positions 40 cm apart. The power of the The distance between an object and the screen is 100 cm. A lens produces an image on the screen [...]