Question

The load arm of a lever is 25 cm long and its effort arm is 75cm long .How much load can be lift by an effort of 4N

Answer 1

Given

• Load arm of a lever = 25 cm
• Length of the effort arm = 75 cm
• Effort = 4N

To Find

• Load that can be lifted

Solution

☯ Effort arm/Load arm = Load/Effort

• We shall use this formula, substitute the values and on simple rearranging of terms we will be able to get our answer!!

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✭ According to the Question :

➞ Effort arm/Load arm = Load/Effort

• Effort Arm = 75 cm
• Load Arm = 25 cm
• Effort = 4 N
• Load = ?

➞ 75/25=Load/4

➞ 300/25 = Load

➞ Load = 12 N

∴ The list that can be lifted with the given effort is 12 N

Answer 2

${\large{\rm{\underline{Let's \; understand \; the \; question}}}}$

$\; \; \;{\bullet}$ This question says that the load arm of a lever is 25 cm in length and it's effort arm is 75 cm in length. We have to find that how much load can be lift by an effort of 4N ?

${\large{\rm{\underline{Given \; that}}}}$

$\; \; \;{\bullet}$ The load arm of a lever is 25 cm in length

$\; \; \;{\bullet}$ The effort arm of a lever is 75 cm in length

$\; \; \;{\bullet}$ Effort is 4N

${\large{\rm{\underline{To \; find}}}}$

$\; \; \;{\bullet}$ How much load can be lift ?

${\large{\rm{\underline{Solution}}}}$

$\; \; \;{\bullet}$ 12N of load can be lift by an effort of 4 Newton.

${\large{\rm{\underline{Using \; formula}}}}$

${\sf{\boxed{\boxed{\longrightarrow \dfrac{Effort \: arm}{Load \: arm} = \dfrac{Load}{Effort}}}}}$

${\large{\rm{\underline{Solution}}}}$

${\sf{\longmapsto \dfrac{Effort \: arm}{Load \: arm} = \dfrac{Load}{Effort}}}$

Here,

⚔️ Effort arm = 75 cm

⚔️ Load arm = 25 cm

⚔️ Effort = 4N

⚔️ Load = [To find]

${\sf{\longmapsto \dfrac{75}{25} = \dfrac{Load}{4}}}$

${\sf{\longmapsto \dfrac{15}{5} = \dfrac{Load}{4}}}$

${\sf{\longmapsto 3 = \dfrac{Load}{4}}}$

${\sf{\longmapsto 3 \times 4 = \: Load}}$

${\sf{\longmapsto 12 = \: Load}}$

${\sf{\longmapsto Load \: = 12N}}$

${\small{\pink{\frak{Henceforth, \: 12N \: is \: the \: load \: can \: be\: lift \: by \: an \: effort \: of \: 4N}}}}$

${\large{\rm{\underline{Additional \: knowledge}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Number \: of \: SI \: units \: are \: 7}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Ampere \: is \: the \: unit \: of \: current \: electricity}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: Young's \: modulus \: of \: elasticity \: is \: Newton/m^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: pressure \: is \: Pascal}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Curie \: is \: the \: unit \: of \: radio \: activity}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Decibel \: is \: the \: unit \: of \: intensity \: of \: sound}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: electric \: charge \: is \: coulomb}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: resistance \: is \: ohm}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: acceleration \: is \: ms^{-2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Maxwell \: is \: unit \: of \: magnetic \: flux}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: magnetic \: flux \: is \: Weber}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: surface \: tension \: is \: \dfrac{N}{m}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: mechanical \: power \: is \: Watt}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Kinetic \: energy \: is \: given \: by \: \dfrac{1}{2}mv^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Value \: of \: G \: is \: 6.673 \times 10^{-11}Nm^{2}kg{-3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Dimensional \: formula \: for \: universal \: gravitational \: constant \: is \: M^{-1} L^{3} T^{-2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto The \: unit \: of \: force \: constant \: k \: of \: a \: spring \: is \: \dfrac{N}{m}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Sir \: Cavendish \: was \: the \: first \: to \: gave \: value \: of \: G \: experimentally}}}$